A165943 - OEIS

%I #32 Oct 08 2024 06:24:34

%S 0,77,63,525,56,1365,483,2597,210,4221,1295,6237,462,8645,2499,11445,

%T 812,14637,4095,18221,1260,22197,6083,26565,1806,31325,8463,36477,

%U 2450,42021,11235,47957,3192,54285,14399,61005,4032,68117,17955,75621,4970

%N Heptasection A061037(7*n+2).

%C The (2k+1)-sections A061037((2*k+1)*n+2) are multiples of 2k+1:

%C 0,...21,...15,..117,...12,..285,...99,..525,...42,..837,..255, k=1, A142590

%C 0,...45,...35,..285,...30,..725,..255,.1365,..110,.2205,..675, k=2, A165248

%C 0,...77,...63,..525,...56,.1365,..483,.2597,..210,.4221,.1295, k=3, here

%C 0,..117,...99,..837,...90,.2205,..783,.4221,..342,.6885,.2115, k=4,

%C 0,..165,..143,.1221,..132,.3245,.1155,.6237,..506,10197,.3135, k=5

%C 0,..221,..195,.1677,..182,.4485,.1599,.8645,..702,14157,.4355, k=6

%C After division by 2k+1 these define a table T'(k,c) :

%C 0,....7,....5,...39,....4,...95,...33,..175,...14,..279,...85, k=1, A142883

%C 0,....9,....7,...57,....6,..145,...51,..273,...22,..441,..135, k=2

%C 0,...11,....9,...75,....8,..195,...69,..371,...30,..603,..185, k=3

%C 0,...13,...11,...93,...10,..245,...87,..469,...38,..765,..235, k=4

%C 0,...15,...13,..111,...12,..295,..105,..567,...46,..927,..285, k=5

%C 0,...17,...15,..129,...14,..345,..123,..665,...54,.1089,..335, k=6

%C Differences downwards each second column in this second table are 2 = 7-5 = 9-7..; 18 = 57-39 = 75-57..; 50 = 145-95 = 195-145... = A077591(n+1) = 2*A016754.

%C The difference T'(k+1,c)-T'(k,c) is 0, 2, 2, 18, 2, 50, 18, 98, 8 ... = 2*A181318(c) =A061037(c-2)+A061037(c+2). - _Paul Curtz_, Mar 12 2012

%C Let b(n)= a(n) mod 11. The sequence b(n) has the property b(n+44) = b(n) with the first 43 values being {0, 0 , 8, 1, 1, 10, 1, 1, 8, 8, 0, 0, 10, 5, 5, 9, 7, 1, 5, 6, 10, 7, 0, 2, 1, 1, 8, 1, 5, 8, 2, 0, 8, 10, 6, 5, 10, 7, 9, 5, 0, 10, 0}. - _G. C. Greubel_, Apr 18 2016

%H G. C. Greubel, <a href="/A165943/b165943.txt">Table of n, a(n) for n = 0..5000</a>

%F a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>12. - Conjectured by _R. J. Mathar_, Mar 02 2010, proved by _Robert Israel_, Apr 20 2016

%F From _Ilya Gutkovskiy_, Apr 19 2016: (Start)

%F G.f.: 7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3).

%F a(n) = -7*n*(7*n + 4)*(27*(-1)^n + 6*cos((Pi*n)/2) - 37)/64. (End)

%p seq(numer(1/4 - 1/(7*n+2)^2), n=0..50); # _Robert Israel_, Apr 20 2016

%t Table[Numerator[1/4 - 1/(7 n + 2)^2], {n, 0, 40}] (* _Michael De Vlieger_, Apr 19 2016 *)

%t CoefficientList[Series[7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3), {x, 0, 50}], x] (* _G. C. Greubel_, Sep 19 2018 *)

%o (PARI) a(n) = numerator(1/4 - 1/(7*n+2)^2); \\  _Altug Alkan_, Apr 18 2016

%o (PARI) x='x+O('x^50); concat([0], Vec(7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))) \\ _G. C. Greubel_, Sep 19 2018

%o (Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))); // _G. C. Greubel_, Sep 19 2018

%K nonn,less

%O 0,2

%A _Paul Curtz_, Oct 01 2009

%E Partially edited and extended by _R. J. Mathar_, Mar 02 2010

%E Removed division by 7 in definition and formula - _R. J. Mathar_, Mar 23 2010