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%I #24 Nov 27 2019 09:55:52
%S 5,13,17,25,29,37,41,53,61,73,89,97,101,109,113,125,137,149,157,169,
%T 173,181,193,197,229,233,241,257,269,277,281,289,293,313,317,337,349,
%U 353,373,389,397,401,409,421,433,449,457,461,509,521,541,557,569,577,593
%N Pythagorean prime powers.
%C 1 + sum of the indices of the first two numbers in A001844 that are divisible by n, if 1 + the sum of those indices equals n. - _Mats Granvik_, Oct 16 2007
%C R. J. Turyn proved [Baliga, et al., p. 129, gives the reference] that Williamson Hadamard matrices exist for 4t = 2(p^k + 1), for all primes p such that p == 1 (mod 4). - _L. Edson Jeffery_, Apr 10 2012
%C A024362(a(n)) = 1. - _Reinhard Zumkeller_, Dec 02 2012
%H Ray Chandler, <a href="/A120960/b120960.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from Reinhard Zumkeller)
%H A. Baliga and K. J. Horadam, <a href="http://ajc.maths.uq.edu.au/pdf/11/ocr-ajc-v11-p123.pdf">Cocyclic Hadamard matrices over Z_t X Z^2_2</a>, Australas. J. Combin. 11(1995), 123-134.
%H Eric Weisstein's World of Mathematics, MathWorld, <a href="http://mathworld.wolfram.com/CenteredSquareNumber.html">Centered Square Number</a>.
%e A001844(1) = 5 is divisible by 5, A001844(3) = 25 is divisible by = 5 and 1+3+1=5, so 5 is a member.
%e A001844(2) = 13 is divisible by = 13, A001844(10) = 221 is divisible by = 13 and 2+10+1=13 so 13 is a member.
%o (Excel) Generate the indices with: =if(mod(1+2*row()*(row()+1);4*column()+1)=0;row();") Then sum the first two indices if it equals the column + 1. - _Mats Granvik_, Oct 16 2007
%o (Haskell)
%o import Data.List (elemIndices)
%o a120960 n = a120960_list !! (n-1)
%o a120960_list = map (+ 1) $ elemIndices 1 a024362_list
%o -- _Reinhard Zumkeller_, Dec 02 2012
%Y Cf. Disjoint union of A002144 and A146945.
%Y Cf. A001844, subsequence of A000961.
%Y Cf. A024409, subsequence of A008846.
%K nonn
%O 1,1
%A _Lekraj Beedassy_, Jul 19 2006