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Numbers k such that 24*k + 6 is congruent to 0 (mod 49).
1

%I #31 Sep 08 2022 08:46:21

%S 12,61,110,159,208,257,306,355,404,453,502,551,600,649,698,747,796,

%T 845,894,943,992,1041,1090,1139,1188,1237,1286,1335,1384,1433,1482,

%U 1531,1580,1629,1678,1727,1776,1825,1874,1923,1972,2021,2070,2119,2168,2217,2266,2315,2364

%N Numbers k such that 24*k + 6 is congruent to 0 (mod 49).

%C Equivalently, indices k for which A016813(k) is a multiple of 49. - _Bruno Berselli_, May 10 2018

%C Numbers k such that k == 12 (mod 49). - _Joerg Arndt_, May 11 2018

%H Vincenzo Librandi, <a href="/A304205/b304205.txt">Table of n, a(n) for n = 1..5000</a>

%H Jahgwon Ju, <a href="https://arxiv.org/abs/1805.03434">Universal sums of generalized pentagonal numbers</a>, arXiv:1805.03434 [math.NT], 2018, page 5 (see Case 4-1).

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

%F G.f.: x*(12 + 25*x - 37*x^2)/(1-x)^3.

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

%F a(n) = 49*n - 37. - _Bruno Berselli_, May 11 2018

%t Table[49 n - 37, {n, 1, 50}] (* _Bruno Berselli_, May 11 2018 *)

%o (Magma) [49*n-37: n in [1..50]];

%Y Cf. A016813.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, May 10 2018