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Numbers congruent to 4 or 21 mod 26.
3

%I #33 Sep 09 2022 22:26:09

%S 4,21,30,47,56,73,82,99,108,125,134,151,160,177,186,203,212,229,238,

%T 255,264,281,290,307,316,333,342,359,368,385,394,411,420,437,446,463,

%U 472,489,498,515,524,541,550,567,576,593,602,619,628,645,654,671,680,697,706,723,732,749,758,775,784,801,810,827,836,853,862

%N Numbers congruent to 4 or 21 mod 26.

%C A007310(a(n)+1) is always a multiple of 13.

%C a(n) mod 6 follows the following pattern: 4,3,0,5,2,1,4,3,0,5,2,1 and so on.

%C a(n) mod 4 = A010873(n)

%C A020639(A007310(a(n)+1) = 5 when n is congruent to 2 or 9 (mod 10) (n is a term in A273669). It equals 7 when n is congruent to 3 or 12 (mod 14) but not congruent to 2 or 9 (mod 10). It equals 11 when n is congruent to 4 or 19 (mod 22) but not congruent to 2 or 9 (mod 10) and not congruent to 3 or 12 (mod 14). Otherwise, it is 13.

%H Davis Smith, <a href="/A306285/b306285.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F a(n) = 13*n - A010720(n+1).

%F From _Colin Barker_, Feb 08 2019: (Start)

%F G.f.: x*(4 + 17*x + 5*x^2) / ((1 - x)^2*(1 + x)).

%F a(n) = 13*n - 5 for n even.

%F a(n) = 13*n - 9 for n odd.

%F a(n) = a(n-1) + a(n-2) - a(n-3) for n>3. (End)

%F E.g.f.: 5 + (13*x - 7)*exp(x) + 2*exp(-x). - _David Lovler_, Sep 09 2022

%p seq(seq(26*i+j, j=[4, 21]), i=0..200);

%t Select[Range[200], MemberQ[{4, 21}, Mod[#, 26]] &]

%o (PARI) for(n=1, 1000, if((n%26==4) || (n%26==21), print1(n, ", ")))

%o (PARI) Vec(x*(4 + 17*x + 5*x^2) / ((1 - x)^2*(1 + x)) + O(x^40)) \\ _Colin Barker_, Feb 08 2019

%Y Cf. A007310, A010720, A020639, A010873, A273669.

%K nonn,easy

%O 1,1

%A _Davis Smith_, Feb 03 2019