A073412 -id:A073412 - cp's OEIS frontend

A064715 Smallest member of triple of consecutive numbers each of which is the sum of two different nonzero squares.

232, 520, 584, 800, 808, 1096, 1224, 1312, 1600, 1664, 1744, 1800, 1872, 1960, 2248, 2312, 2384, 2600, 2824, 3328, 3392, 3600, 4112, 4176, 4328, 4624, 5120, 5328, 5408, 5904, 6056, 6120, 6352, 6408, 6568, 6920, 8080, 8144, 8296, 8352, 8584, 9160, 9376

Offset: 1

Robert G. Wilson v, Oct 13 2001

nonn

All terms == 0 mod 8. Is this the same as A073412? - Zak Seidov, Jan 26 2013

This sequence is distinct from A073412 since it does not allow numbers equal to twice a square, like 72, 1152, 2592, 3528, etc. - Giovanni Resta, Jan 29 2013

			232 = 6^2 + 14^2, 233 = 8^2 + 13^2, and 234 = 3^2 + 15^2.

David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, Penguin Books Ltd., Middlesex, England, 1997, page 133. - "It is not possible to have 4 such consecutive numbers."

Zak Seidov, Table of n, a(n) for n = 1..1200

Cf. A004431.

a = Table[n^2, {n, 1, 100} ]; c = {}; Do[ c = Append[c, a[[i]] + a[[j]]], {i, 1, 100}, {j, 1, i - 1} ]; c = Union[c]; c[[ Select[ Range[ Length[c] - 2], c[[ # ]] + 2 == c[[ # + 2 ]] & ]]]

A274555 Nonsquare n such that n*(n+1)/2 is the sum of two nonzero squares.

Original entry on oeis.org

17, 40, 52, 72, 73, 80, 89, 97, 116, 136, 145, 148, 180, 193, 232, 233, 241, 244, 260, 288, 292, 305, 313, 337, 360, 369, 388, 404, 409, 424, 449, 457, 481, 520, 521, 544, 548, 577, 584, 585, 592, 612, 628, 640, 656, 673, 697, 724, 745, 772, 793, 800

Offset: 1

Altug Alkan, Jul 06 2016

Sequence focuses on the nonsquare numbers in order to eliminate trivial solutions (A000217(n^2) = (n^2 + n^4)/2 = ((n + n^2)/2)^2 + ((n - n^2)/2)^2).
A073412 is a subsequence. Additionally, (A073412(n), A073412(n) + 1) gives consecutive pairs of this sequence that are (72, 73), (232, 233), (520, 521), (584, 585), ...
Proof:
Note that (a^2 + b^2)*(c^2 + d^2)/2 = ((a*c + b*d)^2 + (a*d - b*c)^2)/2 = ((a*c + b*d + a*d - b*c)/2)^2 + ((a*c + b*d - a*d + b*c)/2)^2 and A073412(n) + k is not a square by definition of it for 0 <= k <= 2. So this explains the reason of the fact that A073412(n) and A073412(n) + 1 are always members of this sequence. Furthermore, if there is a consecutive pair in this sequence, the lesser of pair must be in A073412 since n, n+1 and n+2 must be the sum of two nonzero squares, if n*(n+1)/2 and (n+1)*(n+2)/2 are the sum of two nonzero squares and n, n+1, n+2 are nonsquares. So exactly (A073412(n), A073412(n) + 1) gives consecutive pairs of this sequence.

			17 is a term because 17 is not a square and 17*(17+1)/2 = 153 = 3^2 + 12^2.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A000037, A000217, A000404, A073412, A274686.

isA000404(n) = for( i=1, #n=factor(n)~%4, n[1, i]==3 && n[2, i]%2 && return); n && ( vecmin(n[1, ])==1 || (n[1, 1]==2 && n[2, 1]%2));
lista(nn) = for(n=1, nn, if(!issquare(n) && isA000404(n*(n+1)/2), print1(n, ", ")));

has(n)=my(f=factor(n)); for(i=1,#f~, if(f[i,1]%4>2 && f[i,2]%2, return(0))); if(#select(p->p%4==1, f[,1]), 2, 1)
is(n)=my(t); if(n%4>1 || issquare(n), return(0)); t=has(numerator(n/2)); t && if(t>1, has(numerator((n+1)/2)), t=has(numerator((n+1)/2)); t && (valuation(n*(n+1),2)%2==0 || t>1)) \\ Charles R Greathouse IV, Jul 19 2016

cp's OEIS Frontend

A064715 Smallest member of triple of consecutive numbers each of which is the sum of two different nonzero squares.

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