A218395 -id:A218395 - cp's OEIS frontend

A257825 Positive integers whose square is the sum of 74 consecutive squares.

2257, 2849, 21941, 27713, 604765, 763865, 16669573, 21054961, 162316669, 205018517, 4474051285, 5651073085, 123321498797, 155764598629, 1200818695321, 1516726961053, 33099030801665, 41806637918965, 912332431430633, 1152346479602381, 8883656545668089

Offset: 1

Colin Barker, May 10 2015

Positive integers x in the solutions to 2*x^2-148*y^2-10804*y-264698 = 0.

			2257 is in the sequence because 2257^2 = 5094049 = 225^2+226^2+...+298^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,7398,0,0,0,0,0,-1).

Cf. A001653, A180274, A218395, A257761, A257765, A257767, A257780, A257781, A257823, A257824, A257826-A257828.

I:=[2257,2849,21941,27713,604765,763865,16669573, 21054961,162316669,205018517,4474051285,5651073085]; [n le 12 select I[n] else 7398*Self(n-6)-Self(n-12): n in [1..40]]; // Vincenzo Librandi, May 11 2015

 LinearRecurrence[{0, 0, 0, 0, 0, 7398, 0, 0, 0, 0, 0, -1}, {2257, 2849, 21941, 27713, 604765, 763865, 16669573, 21054961, 162316669, 205018517, 4474051285, 5651073085}, 40] (* Vincenzo Librandi, May 11 2015 *)

Vec(-37*x*(5*x^11+5*x^10+61*x^9+77*x^8+593*x^7+749*x^6-20645*x^5-16345*x^4-749*x^3-593*x^2-77*x-61) / ((x^6-86*x^3-1)*(x^6+86*x^3-1)) + O(x^100))

a(n) = 7398*a(n-6)-a(n-12).

G.f.: -37*x*(5*x^11+5*x^10+61*x^9+77*x^8+593*x^7+749*x^6-20645*x^5-16345*x^4-749*x^3-593*x^2-77*x-61) / ((x^6-86*x^3-1)*(x^6+86*x^3-1)).

A257827 Positive integers whose square is the sum of 96 consecutive squares.

Original entry on oeis.org

652, 724, 788, 1012, 1828, 2372, 2596, 2908, 6164, 6908, 7564, 9836, 17996, 23404, 25628, 28724, 60988, 68356, 74852, 97348, 178132, 231668, 253684, 284332, 603716, 676652, 740956, 963644, 1763324, 2293276, 2511212, 2814596, 5976172, 6698164, 7334708

Offset: 1

Colin Barker, May 10 2015

Positive integers x in the solutions to 2*x^2-192*y^2-18240*y-580640 = 0.

			652 is in the sequence because 652^2 = 425104 = 13^2+14^2+...+108^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,10,0,0,0,0,0,0,0,-1).

Cf. A001653, A180274, A218395, A257761, A257765, A257767, A257780, A257781, A257823-A257826, A257828.

I:=[652,724,788,1012,1828,2372,2596,2908,6164,6908, 7564,9836,17996,23404,25628,28724]; [n le 16 select I[n] else 10*Self(n-8)-Self(n-16): n in [1..40]]; // Vincenzo Librandi, May 11 2015

LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, -1}, {652, 724, 788, 1012, 1828, 2372, 2596, 2908, 6164, 6908, 7564, 9836, 17996, 23404, 25628, 28724}, 40] (* Vincenzo Librandi, May 11 2015 *)

Vec(-4*x*(89*x^15 +83*x^14 +79*x^13 +71*x^12 +71*x^11 +79*x^10 +83*x^9 +89*x^8 -727*x^7 -649*x^6 -593*x^5 -457*x^4 -253*x^3 -197*x^2 -181*x -163) / (x^16-10*x^8+1) + O(x^100))

a(n) = 10*a(n-8) -a(n-16).

G.f.: -4*x*(89*x^15 +83*x^14 +79*x^13 +71*x^12 +71*x^11 +79*x^10 +83*x^9 +89*x^8 -727*x^7 -649*x^6 -593*x^5 -457*x^4 -253*x^3 -197*x^2 -181*x-163) / (x^16 -10*x^8 +1).

A201632 If the sum of the squares of 4 consecutive numbers is a triangular number t(u), then a(n) is its index u.

Original entry on oeis.org

35, 83, 1203, 2835, 40883, 96323, 1388835, 3272163, 47179523, 111157235, 1602714963, 3776073843, 54445129235, 128275353443, 1849531679043, 4357585943235, 62829631958243, 148029646716563, 2134357954901235, 5028650402419923

Offset: 1

Paul Weisenhorn, Jan 09 2013

Sum_{(e(n)+j)^2,j=0..3} = a(n)*(a(n)+1)/2=t(a(n)) give the Pell equation c(n)^2 - 32*d(n)^2 = 41 with 2*a(n) + 1 = c(n) and e(n) + 1.5 = d(n). e(n) = A201633(n).

In general, for the sum of the squares of k consecutive numbers, one get an analog sequence with k in {4, 5, 6, 7, 11, 15, 17, 19, 23,...}. It gives the Pell equation c(n)^2 - 8k*d(n)^2 = 4*binomial((k+1),3) + 1 with 2*a(n) + 1 = c(n) and e(n) + (k-1)/2 = d(n).

			For n=2: a(2)=83; t(83)=83*84/2=3486.
A201633(2)=e(2)=28; 28^2+29^2+30^2+31^2=3486.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A201633, A218395.

LinearRecurrence[{1,34,-34,-1,1},{35,83,1203,2835,40883},30] (* Harvey P. Dale, Dec 10 2024 *)

G.f.: (35*x+48*x^2-70*x^3+3*x^5)/((1-x)*(1-34*x^2+x^4)).
a(n+4) = 34*a(n+2) - a(n) + 16.
a(n+5) = a(n+4) + 34*a(n+3) - 34*a(n+2) - a(n+1) + a(n).
eigenvalues ej: {1,(3+2r),-(3+2r),(3-2r),-(3-2r)}.
a(n+1) = (k1*e1 + k2*e2^n + k3*e3^n + k4*e4^n + k5*e5^n)/4 for k1=-2; k2=50+35r; k3=21+15r; k4=50-35r; k5=21-15r, where r = sqrt(2).

Corrected by R. J. Mathar, Jun 14 2016

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A257825 Positive integers whose square is the sum of 74 consecutive squares.

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A257827 Positive integers whose square is the sum of 96 consecutive squares.

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A201632 If the sum of the squares of 4 consecutive numbers is a triangular number t(u), then a(n) is its index u.

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