A094196 -id:A094196 - cp's OEIS frontend

A180274 Integers whose squares are the sums of 24 consecutive squares.

70, 106, 158, 182, 274, 430, 650, 1022, 1546, 1786, 2702, 4250, 6430, 10114, 15302, 17678, 26746, 42070, 63650, 100118, 151474, 174994, 264758, 416450, 630070, 991066, 1499438, 1732262, 2620834, 4122430, 6237050, 9810542, 14842906, 17147626, 25943582

Offset: 1

Zhining Yang, Jan 17 2011

The corresponding starts of 24 consecutive squares to be summed are A094196.

Colin Barker, Table of n, a(n) for n = 1..1000
K. S. Brown, Sum of Consecutive Nth Powers Equals an Nth Power
Fun With Num3ers, Sets of 24 consecutive squares whose sum is a square, July 27 2016.
V. Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014. See Table 2 p. 8.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,10,0,0,0,0,0,-1).

Cf. A094196.

Cf. A001032 (24 is a term of that sequence).

A094196 := proc(n) if n <= 12 then op(n,[1, 9, 20, 25, 44, 76, 121, 197, 304, 353, 540, 856]) ; else 10*procname(n-6)-procname(n-12)+92 ; end if ; end proc:
A180274 := proc(n) local a96 ; a96 := A094196(n) ; 24*a96^2+552*a96+4324 ; sqrt(%) ; end proc:
seq(A180274(n),n=1..30) ; # R. J. Mathar, Jan 20 2011

Select[Sqrt[#]&/@(Total[#]&/@Partition[Range[900000]^2, 24, 1]), IntegerQ] (* Harvey P. Dale, Jan 21 2011 *)
t={70, 106, 158, 182, 274, 430, 650, 1022, 1546, 1786, 2702, 4250}; Do[AppendTo[t, 10*t[[-6]] - t[[-12]]], {n, 13, 100}]; t

{ for(n=1,999999,t=((n+23)*(n+24)*(2*n+47)-n*(n-1)*(2*n-1))/6;if(issquare(t),print1(ceil(sqrt(t)),","))) }

Vec(-2*x*(25*x^11+19*x^10+17*x^9+17*x^8+19*x^7+25*x^6-215*x^5-137*x^4-91*x^3-79*x^2-53*x-35) / (x^12-10*x^6+1) + O(x^100)) \\ Colin Barker, May 09 2015

a(n) = +10*a(n-6) -a(n-12). G.f. ( 70+106*x+158*x^2+182*x^3+274*x^4+430*x^5-50*x^6-38*x^7-34*x^8-34*x^9-38*x^10-50*x^11 ) / ( 1-10*x^6+x^12 ). - Joerg Arndt, Jan 17 2011

a(n) = sqrt( 24*(A094196(n))^2 +552*A094196(n)+4324) . - R. J. Mathar, Jan 20 2011

A106521 Numbers m such that Sum_{k=0..10} (m+k)^2 is a square.

Original entry on oeis.org

18, 38, 456, 854, 9192, 17132, 183474, 341876, 3660378, 6820478, 73024176, 136067774, 1456823232, 2714535092, 29063440554, 54154634156, 579811987938, 1080378148118, 11567176318296, 21553408328294

Offset: 1

Ralf Stephan, May 30 2005

Equivalently, 11*a(n)^2 + 110*a(n) + 385 is a square.
11*((m+5)^2+10) is a square iff the second factor is divisible by 11 and the quotient is a square, i.e., iff m = 11*k - 4 or m = 11*k - 6 and 11*k^2 +- 2 k + 1 is a square. Thus a(n) == (7,5,5,7,7,5,5,7,...) (mod 11), repeating with period 4 and the values are obtained by solving these Pell-type equations (cf. link to Dario Alpern's quadratic solver). The corresponding recurrence equations (see PARI code) should make it possible to prove the conjectured g.f. - M. F. Hasler, Jan 27 2008
All sequences of this type (i.e., sequences with fixed offset k, and a discernible pattern: k=0...10 for this sequence, k=0..1 for A001652) can be continued using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 05 2016

			Since 18^2 + 19^2 + ... + 28^2 = 5929 = 77^2, 18 is in the sequence. - _Michael B. Porter_, Aug 07 2016

Vincenzo Librandi, Table of n, a(n) for n = 1..500
Dario Alpern, Quadratic two integer variable equation solver
Index entries for linear recurrences with constant coefficients, signature (1,20,-20,-1,1).

Cf. A001032, A094196.

LinearRecurrence[{1,20,-20,-1,1},{18,38,456,854,9192},30] (* Harvey P. Dale, May 07 2011 *)

A106521(n)={local(xy=[ -4-2*(n%2);11],PQRS=[10,3;33,10],KL=[45;165]);until(0>=n-=2,xy=PQRS*xy+KL);xy[1]} \\ M. F. Hasler, Jan 27 2008

G.f.: 2*x*(9+10*x+29*x^2-x^3-2*x^4)/(1-x)/(1-20*x^2+x^4). - Vladeta Jovovic, May 31 2005; adapted to the offset by Bruno Berselli, May 16 2011
a(1)=18, a(2)=38, a(3)=456, a(4)=854, a(5)=9192; thereafter a(n)=a(n-1)+20*a(n-2)- 20*a(n-3)-a(n-4)+a(n-5). - Harvey P. Dale, May 07 2011
a(n) = A198949(n+1)-5. - Bruno Berselli, Feb 12 2012
a(1)=18, a(2)=38, a(3)=456, a(4)=854; thereafter a(n) = 20*a(n-2) - a(n-4) + 90. - Daniel Mondot, Aug 05 2016

Edited and extended by M. F. Hasler, Jan 27 2008

A269447 The first of 23 consecutive positive integers the sum of the squares of which is a square.

Original entry on oeis.org

7, 17, 881, 1351, 42787, 65337, 2053401, 3135331, 98520967, 150431057, 4726953521, 7217555911, 226795248547, 346292253177, 10881444977241, 16614810597091, 522082563659527, 797164616407697, 25049081610680561, 38247286776972871, 1201833834749007907

Offset: 1

Colin Barker, Feb 27 2016

Positive integers y in the solutions to 2*x^2-46*y^2-1012*y-7590 = 0.

All sequences of this type (i.e. sequences with fixed offset k, and a discernible pattern: k=0...22 for this sequence, k=0..1 for A001652, k=0...10 for A106521) can be continued using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 05 2016

			7 is in the sequence because sum(k=7, 29, k^2) = 8464 = 92^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,48,-48,-1,1).

Cf. A001032, A001652, A094196, A106521, A257761, A269448, A269449, A269451.

LinearRecurrence[{1,48,-48,-1,1},{7,17,881,1351,42787},30] (* Harvey P. Dale, May 21 2024 *)

Vec(x*(7+10*x+528*x^2-10*x^3-29*x^4)/((1-x)*(1-48*x^2+x^4)) + O(x^30))

a(n) = a(n-1)+48*a(n-2)-48*a(n-3)-a(n-4)+a(n-5) for n>5.
G.f.: x*(7+10*x+528*x^2-10*x^3-29*x^4) / ((1-x)*(1-48*x^2+x^4)).
a(1)=7, a(2)=17, a(3)=881, a(4)=1351, a(n) = 48*a(n-2)-a(n-4)+506. - Daniel Mondot, Aug 05 2016

A269448 The first of 26 consecutive positive integers the sum of the squares of which is a square.

Original entry on oeis.org

25, 301, 454, 3850, 31966, 47569, 393925, 3261481, 4852834, 40177750, 332640346, 494942749, 4097737825, 33926055061, 50479308814, 417929081650, 3460124977126, 5148394557529, 42624668591725, 352898821613041, 525085765560394, 4347298267275550

Offset: 1

Colin Barker, Feb 27 2016

Positive integers y in the solutions to 2*x^2-52*y^2-1300*y-11050 = 0.

All sequences of this type (i.e. sequences with fixed offset k, and a discernible pattern: k=0...25 for this sequence, k=0...22 for A269447, k=0..1 for A001652) can be continued using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 05 2016

			25 is in the sequence because sum(k=25, 50, k^2) = 38025 = 195^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,102,-102,0,-1,1).

Cf. A001032, A001652, A094196, A106521, A257765, A269447, A269449, A269451.

Rest@ CoefficientList[Series[x (25 + 276 x + 153 x^2 + 846 x^3 - 36 x^4 - 3 x^5 - 11 x^6)/((1 - x) (1 - 102 x^3 + x^6)), {x, 0, 22}], x] (* Michael De Vlieger, Aug 07 2016 *)

Vec(x*(25+276*x+153*x^2+846*x^3-36*x^4-3*x^5-11*x^6)/((1-x)*(1-102*x^3+x^6)) + O(x^30))

G.f.: x*(25+276*x+153*x^2+846*x^3-36*x^4-3*x^5-11*x^6) / ((1-x)*(1-102*x^3+x^6)).

a(1)=25, a(2)=301, a(3)=454, a(4)=3850, a(5)=31966, a(6)=47569, a(n)=102*a(n-3) - a(n-6) + 1250. - Daniel Mondot, Aug 05 2016

A269449 The first of 33 consecutive positive integers the sum of the squares of which is a square.

Original entry on oeis.org

7, 27, 60, 181, 227, 612, 1085, 1985, 3492, 9047, 11161, 28860, 50607, 91987, 161276, 416685, 513883, 1327652, 2327541, 4230121, 7415908, 19159167, 23628161, 61043836, 107016983, 194494283, 340971196, 880905701, 1086382227, 2806689508, 4920454381, 8942507601

Offset: 1

Colin Barker, Feb 27 2016

Positive integers y in the solutions to 2*x^2-66*y^2-2112*y-22880 = 0.
All sequences of this type (i.e. sequences with fixed offset k, and a discernible pattern: k=0...32 for this sequence, k=0..1 for A001652, k=0...10 for A106521) can be extended using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 08 2016
Numbers x such that 11440+33*x*(32+x)is a square. - Harvey P. Dale, Oct 18 2020

			7 is in the sequence because sum(k=7, 39, k^2) = 20449 = 143^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,46,-46,0,0,0,0,-1,1).

Cf. A001032, A001652, A094196, A106521, A257767, A269447, A269448, A269451.

Rest@ CoefficientList[Series[x (7 + 20 x + 33 x^2 + 121 x^3 + 46 x^4 + 385 x^5 + 151 x^6 - 20 x^7 - 11 x^8 - 11 x^9 - 2 x^10 - 11 x^11 - 4 x^12)/((1 - x) (1 - 46 x^6 + x^12)), {x, 0, 32}], x] (* Michael De Vlieger, Aug 08 2016 *)
LinearRecurrence[{1,0,0,0,0,46,-46,0,0,0,0,-1,1},{7,27,60,181,227,612,1085,1985,3492,9047,11161,28860,50607},50] (* Harvey P. Dale, Oct 18 2020 *)

Vec(x*(7 +20*x +33*x^2 +121*x^3 +46*x^4 +385*x^5 +151*x^6 -20*x^7 -11*x^8 -11*x^9 -2*x^10 -11*x^11 -4*x^12) / ((1 -x)*(1 -46*x^6 +x^12)) + O(x^40))

G.f.: x*(7 +20*x +33*x^2 +121*x^3 +46*x^4 +385*x^5 +151*x^6 -20*x^7 -11*x^8 -11*x^9 -2*x^10 -11*x^11 -4*x^12) / ((1 -x)*(1 -46*x^6 +x^12)).

a(1)=7, a(2)=27, a(3)=60, a(4)=181, a(5)=227, a(6)=612, a(7)=1085, a(8)=1985, a(9)=3492, a(10)=9047, a(11)=11161, a(12)=28860, a(n)=46*a(n-6)-a(n-12)+704. - Daniel Mondot, Aug 08 2016

A269451 The first of 50 consecutive positive integers the sum of the squares of which is a square.

Original entry on oeis.org

7, 28, 44, 67, 87, 124, 168, 287, 379, 512, 628, 843, 1099, 1792, 2328, 3103, 3779, 5032, 6524, 10563, 13687, 18204, 22144, 29447, 38143, 61684, 79892, 106219, 129183, 171748, 222432, 359639, 465763, 619208, 753052, 1001139, 1296547, 2096248, 2714784

Offset: 1

Colin Barker, Feb 27 2016

Positive integers y in the solutions to 2*x^2-100*y^2-4900*y-80850 = 0.

Numbers n such that 40425 + 2450*n + 50*n^2 is a square. - Harvey P. Dale, Oct 22 2016

			7 is in the sequence because sum(k=7, 56, k^2) = 60025 = 245^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,6,-6,0,0,0,0,-1,1).

Cf. A001032, A001652, A094196, A106521, A257781, A269447, A269448, A269449.

Select[Range[3*10^6],IntegerQ[Sqrt[40425+2450#+50#^2]]&] (* or *) LinearRecurrence[ {1,0,0,0,0,6,-6,0,0,0,0,-1,1},{7,28,44,67,87,124,168,287,379,512,628,843,1099},40] (* Harvey P. Dale, Oct 22 2016 *)

Vec(x*(7+21*x+16*x^2+23*x^3+20*x^4+37*x^5+2*x^6-7*x^7-4*x^8-5*x^9-4*x^10-7*x^11-x^12) / ((1-x)*(1+2*x^3-x^6)*(1-2*x^3-x^6)) + O(x^40))

G.f.: x*(7+21*x+16*x^2+23*x^3+20*x^4+37*x^5+2*x^6-7*x^7-4*x^8-5*x^9-4*x^10-7*x^11-x^12) / ((1-x)*(1+2*x^3-x^6)*(1-2*x^3-x^6)).

A180259 Squares which are the sum of consecutive squares starting with 25^2.

Original entry on oeis.org

625, 33124, 38025, 127449, 64529089, 81180100, 15884821225, 3370675683600

Offset: 1

Zhining Yang, Jan 17 2011

That is, terms are squares of the form sum_{i=25..m} i^2 = (m-24) *(2*m^2+51*m+1225) / 6 for some m. Known solutions refer to m = 25, 48, 50, 73, 578, 624, 3625 and 21624, and no further in the range m <= 70000000.

This sequence is complete. See A180442 and A184763.

			38025 is in the sequence because 38025 = 195^2 = 25^2 + 26^2 + ... + 50^2.

Cf. A001032, A094196, A151557, A180273.

Select[Accumulate[Range[25, 22000]^2], IntegerQ[Sqrt[#]] &] (* Harvey P. Dale, Aug 10 2023 *)

for(n=26,9999999,t=n*(n+1)*(2*n+1)/6-4900;if(issquare(t),print1(t,",")))

A180273 Squares which are a sum of consecutive squares starting with 7^2.

Original entry on oeis.org

49, 8464, 20449, 60025, 2304324, 3624882849

Offset: 1

Zhining Yang, Jan 17 2011

That is, terms are squares of the form sum_{i=7..m} i^2 for some m. This sequence is complete. See A180442 and A184763.

			a(2) = 8464 = 92^2 = 7^2+8^2+9^2+...+28^2+29^2.

Cf. A001032, A094196, A151557, A180259.

Select[Accumulate[Range[7,3000]^2],IntegerQ[Sqrt[#]]&] (* Harvey P. Dale, Jul 16 2014 *)

for(n=8,9999999,t=n*(n+1)*(2*n+1)/6-91;if(issquare(t),print1(t,",")))

A180465 Squares which are a sum of consecutive squares starting with 38^2.

Original entry on oeis.org

1444, 20449, 281961, 14212900, 107827456, 564343507984

Offset: 1

Zhining Yang, Jan 19 2011

This sequence is complete. See A180442 and A184763.

Cf. A001032, A094196, A151557, A180259, A180273

for(n=38, 9999999, t=n*(n+1)*(2*n+1)/6-17575; if(issquare(t), print1(t, ", ")))

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A180274 Integers whose squares are the sums of 24 consecutive squares.

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A106521 Numbers m such that Sum_{k=0..10} (m+k)^2 is a square.

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A269447 The first of 23 consecutive positive integers the sum of the squares of which is a square.

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A269448 The first of 26 consecutive positive integers the sum of the squares of which is a square.

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A269449 The first of 33 consecutive positive integers the sum of the squares of which is a square.

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A269451 The first of 50 consecutive positive integers the sum of the squares of which is a square.

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A180259 Squares which are the sum of consecutive squares starting with 25^2.

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