proposed

approved

editing

proposed

Cf. A000742, A023022, A023024, A078374, A101271, A339672, A340719, A341870, A341912, A341913, A341914.

allocated for Ilya GutkovskiyNumber of partitions of n into 4 distinct and relatively prime parts.

1, 1, 2, 3, 5, 6, 9, 11, 15, 18, 22, 27, 33, 39, 45, 54, 61, 72, 79, 94, 101, 120, 127, 149, 158, 185, 189, 225, 231, 267, 274, 321, 319, 378, 377, 435, 439, 511, 495, 588, 577, 661, 656, 764, 729, 863, 836, 954, 939, 1089, 1022, 1215, 1165, 1323, 1289, 1492, 1392, 1650, 1566, 1776, 1715

10,3

G.f.: Sum_{k>=1} mu(k)* x^(10*k) / Product_{j=1..4} (1 - x^(j*k)).

nmax = 70; CoefficientList[Series[Sum[MoebiusMu[k] x^(10 k)/Product[1 - x^(j k), {j, 1, 4}], {k, 1, nmax}], {x, 0, nmax}], x] // Drop[#, 10] &

Cf. A000742, A023022, A023024, A078374, A101271.

allocated

nonn

Ilya Gutkovskiy, Feb 23 2021

approved

editing

allocated for Ilya Gutkovskiy

recycled

allocated

editing

approved

Numbers m such that m*(3m+2) is a square.

2, 32, 450, 6272, 87362, 1216800, 16947842, 46302366, 177343944, 206698468, 236052992, 265407516, 279030998, 294762040, 308385522, 324116564, 337740046, 353471088, 367094570, 380718052, 396449094, 425803618, 439427100, 455158142, 470889184, 484512666, 498136148, 511759630, 513867190

1,1

This uses the solution for a k-agon based pyramid, with k=8, for an octagonal base pyramid. From the general formula: Sum = (k/2 - 1)*n^2 + (-k/2 + 2)*n.

a(n) = 2*A098301(n).

a(1) = 2 is in the series as 2*(3*2 + 2) = 16 which is square.

a(2) = 32 is in the series as 2*(3*32 + 2) = 196 which is square.

(MMBasic)

Function sum(x)

k=8 'an octagonal based pyramid

sum = ((0.5*k-1)*x^2)-((-0.5*k+2)*x)

End Function

Function is_square$(i)

If Sqr(i) = Int(Sqr(i)) Then

is_square$="true"

Else

is_square$="false"

End If

End Function

Sub square()

For x = 1 To 10000000

If is_square$(sum(x)) = "true" Then

Print x

Print #1, x

End If

Next x

End Sub

square()

(Python 3)

import math

def sumx(x):

k=8

return ((0.5*k-1)*x**2)-((-0.5*k+2)*x)

def is_square(i):

if math.sqrt(i) == int(math.sqrt(i)):

is_square="true"

else:

is_square="false"

return is_square

for t in range(1000000000):

if is_square(sumx(t)) == "true":

print(t)

Cf. A000290, A098301, A055997(k=3), A220186(k=5), A084703(k=6), A224419(k=7).

nonn,changed

recycled

Glen Gilchrist, Feb 23 2021

proposed

editing

editing

proposed

Cf. A000290, A098301, A055997(k=3), A220186(k=5), A084703(k=6), A224419(k=7).

Numbers n m such that nm*(3n3m+2) is a square.

0,1,1