cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Views

Author

Colin Barker, Feb 27 2016

Keywords

Comments

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

Examples

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

Links

Crossrefs

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

Programs

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

Formula

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

Views

Author

Colin Barker, Feb 27 2016

Keywords

Comments

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

Examples

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

Links

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • PARI
    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))

Formula

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

Views

Author

Colin Barker, Feb 27 2016

Keywords

Comments

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

Examples

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

Links

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • PARI
    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))

Formula

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
Showing 1-3 of 3 results.

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