A076296 -id:A076296 - cp's OEIS frontend

A118673 Positive solutions x to the equation x^2 + (x+71)^2 = y^2.

0, 13, 160, 213, 280, 1113, 1420, 1809, 6660, 8449, 10716, 38989, 49416, 62629, 227416, 288189, 365200, 1325649, 1679860, 2128713, 7726620, 9791113, 12407220, 45034213, 57066960, 72314749, 262478800, 332610789, 421481416, 1529838729, 1938597916, 2456573889

Offset: 0

Mohamed Bouhamida, May 19 2006

Consider all Pythagorean triples (x,x+71,y) ordered by increasing y; sequence gives x values.
For the generic case x^2+(x+p)^2=y^2 with p=2*m^2-1 a prime number in A066436, m>=2 the associated value in A066049, the x values are given by the sequence defined by: a(n) = 6*a(n-3) -a(n-6) + 2*p with a(0)=0, a(1)=2m+1, a(2)=6m^2-10m+4, a(3)=3p, a(4)=6m^2+10m+4, a(5)=40m^2-58m+21.
For the generic case x^2+(x+p)^2=y^2 with p=2*m^2-1 a prime number in A066436, m>=2, Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(0)=p, b(1)=2m^2+2m+1, b(2)=10m^2-14m+5, b(3)=5p, b(4)=10m^2+14m+5, b(5)=58m^2-82m+29. - Mohamed Bouhamida, Sep 09 2009

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

Cf. A076296 (p=7), A118120 (p=17), A118674 (p=31), A129836 (p=97), A129992 (p=127), A129993 (p=199), A129991 (p=241), A129999 (p=337), A130004 (p=449), A130005 (p=577), A130013 (p=647), A130014 (p=881), A130017 (p=967).

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(13+147*x+53*x^2-11*x^3-49*x^4-11*x^5)/((1-x)*(1 - 6*x^3 +x^6)))); // G. C. Greubel, May 07 2018

Select[Range[0,100000],IntegerQ[Sqrt[#^2+(#+71)^2]]&] (* or *) LinearRecurrence[{1,0,6,-6,0,-1,1},{0,13,160,213,280,1113,1420},100] (* Vladimir Joseph Stephan Orlovsky, Feb 02 2012 *)

a(n)=([0,1,0,0,0,0,0; 0,0,1,0,0,0,0; 0,0,0,1,0,0,0; 0,0,0,0,1,0,0; 0,0,0,0,0,1,0; 0,0,0,0,0,0,1; 1,-1,0,-6,6,0,1]^n*[0;13;160;213;280;1113;1420])[1,1] \\ Charles R Greathouse IV, Apr 22 2016

x='x+O('x^30); concat([0], Vec(x*(13+147*x+53*x^2-11*x^3 -49*x^4 -11*x^5)/((1-x)*(1-6*x^3+x^6)))) \\ G. C. Greubel, May 07 2018

a(n) = 6*a(n-3) -a(n-6) +142 with a(0)=0, a(1)=13, a(2)=160, a(3)=213, a(4)=280, a(5)=1113.

O.g.f.: x*(13+147*x+53*x^2-11*x^3-49*x^4-11*x^5)/((1-x)*(1-6*x^3+x^6)). - R. J. Mathar, Jun 10 2008

Edited by R. J. Mathar, Jun 10 2008

A076293 Numbers k where the root mean square (RMS) of k and 7 is an integer, i.e., sqrt((k^2 + 7^2)/2) is an integer.

Original entry on oeis.org

1, 7, 17, 23, 49, 103, 137, 287, 601, 799, 1673, 3503, 4657, 9751, 20417, 27143, 56833, 118999, 158201, 331247, 693577, 922063, 1930649, 4042463, 5374177, 11252647, 23561201, 31322999, 65585233, 137324743, 182563817, 382258751, 800387257, 1064059903

Offset: 0

Henry Bottomley, Oct 05 2002

			17 is in the sequence since sqrt((17^2 + 7^2)/2) = 13 is an integer.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-1).

Cf. A002315, A076294, A076295, A076296.

Column[LinearRecurrence[{0, 0, 6, 0, 0, -1}, {1, 7, 17, 23, 49, 103}, 35] ] (* Vincenzo Librandi, Jul 30 2017 *)

Vec((x+1)*(x^2+3*x+1)^2/(x^6-6*x^3+1) + O(x^100)) \\ Colin Barker, Sep 14 2014

a(n) = 6a(n-3) - a(n-6) = sqrt(2*A076294(n)^2 - 49) = A076295(n) + A076296(n).
a(3n+1) = 7*A002315(n).
G.f.: (x+1)*(x^2+3*x+1)^2 / (x^6-6*x^3+1). - Colin Barker, Sep 14 2014

More terms from Colin Barker, Sep 14 2014

A076294 Consider all Pythagorean triples (X,X+7,Z); sequence gives Z values.

Original entry on oeis.org

5, 7, 13, 17, 35, 73, 97, 203, 425, 565, 1183, 2477, 3293, 6895, 14437, 19193, 40187, 84145, 111865, 234227, 490433, 651997, 1365175, 2858453, 3800117, 7956823, 16660285, 22148705, 46375763, 97103257, 129092113, 270297755, 565959257, 752403973, 1575410767

Offset: 0

Henry Bottomley, Oct 05 2002

First two terms included for consistency with A076293.

For the generic case x^2+(x+p)^2=y^2 with p=2*m^2-1 a prime number in A066436, m>=2, the x values are given by the sequence defined by: a(n)=6*a(n-3)-a(n-6)+2p with a(1)=0, a(2)=2m+1, a(3)=6m^2-10m+4, a(4)=3p, a(5)=6m^2+10m+4, a(6)=40m^2-58m+21.Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=2*m^2+2m+1, b(3)=10m^2-14m+5, b(4)=5p, b(5)=10m^2+14m+5, b(6)=58m^2-82m+29. - Mohamed Bouhamida, Sep 09 2009

			17 is in the sequence as the hypotenuse of the (8,15,17) triangle.

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

Cf. A001653, A076293, A076295, A076296.

LinearRecurrence[{0,0,6,0,0,-1},{5,7,13,17,35,73},40] (* Harvey P. Dale, Mar 19 2019 *)

Vec((1 - x)*(5 + 12*x + 25*x^2 + 12*x^3 + 5*x^4) / (1 - 6*x^3 + x^6) + O(x^50)) \\ Colin Barker, Apr 25 2017

a(n) = 6*a(n-3)-a(n-6) = sqrt((A076293(n)^2+49)/2) = sqrt(A076295(n)^2 + A076296(n)^2).
a(3n+1) = 7*A001653(n).
G.f.: (1 - x)*(5 + 12*x + 25*x^2 + 12*x^3 + 5*x^4) / (1 - 6*x^3 + x^6). - Colin Barker, Apr 25 2017

A076295 Consider all Pythagorean triples (Y-7,Y,Z); sequence gives Y values.

Original entry on oeis.org

4, 7, 12, 15, 28, 55, 72, 147, 304, 403, 840, 1755, 2332, 4879, 10212, 13575, 28420, 59503, 79104, 165627, 346792, 461035, 965328, 2021235, 2687092, 5626327, 11780604, 15661503, 32792620, 68662375, 91281912, 191129379, 400193632, 532029955, 1113983640

Offset: 0

Henry Bottomley, Oct 05 2002

nonn

First two terms included for consistency with A076293.

			15 is in the sequence as the longer leg of the (8,15,17) triangle.

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

Cf. A046090, A076293, A076294, A076296.

LinearRecurrence[{1,0,6,-6,0,-1,1},{4,7,12,15,28,55,72},40] (* Harvey P. Dale, Feb 02 2012 *)

a(n) =6a(n-3)-a(n-6)-14 =(A076293(n)+7)/2 =sqrt(A076294(n)^2-A076296(n)^2) =A076296(n)+7.
a(3n+1) = 7*A046090(n).
a(0)=4, a(1)=7, a(2)=12, a(3)=15, a(4)=28, a(5)=55, a(6)=72, a(n)= a(n-1)+ 6*a(n-3)-6*a(n-4)-a(n-6)+a (n-7). - Harvey P. Dale, Feb 02 2012
G.f.: -(3*x^6-3*x^5-5*x^4-21*x^3+5*x^2+3*x+4) / ((x-1)*(x^6-6*x^3+1)). - Colin Barker, Sep 14 2014

A129010 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+833)^2 = y^2.

Original entry on oeis.org

0, 124, 168, 187, 343, 399, 595, 624, 915, 952, 1260, 1372, 1768, 1827, 1975, 2499, 3135, 3367, 3468, 4312, 4620, 5712, 5875, 7524, 7735, 9499, 10143, 12427, 12768, 13624, 16660, 20352, 21700, 22287, 27195, 28987, 35343, 36292, 45895, 47124

Offset: 1

Mohamed Bouhamida, May 27 2007

nonn

Also values x of Pythagorean triples (x, x+833, y); 833=7^2*17.
Corresponding values y of solutions (x, y) are in A156835.
lim_{n -> infinity} a(n)/a(n-15) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = ((9+4*sqrt(2))/7)^2*((19+6*sqrt(2))/17)/(3+2*sqrt(2)) for n mod 15 = {1, 2, 5, 7, 11, 13}.
lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2))/(((9+4*sqrt(2))/7)*((19+6*sqrt(2))/17)^2) for n mod 15 = {0, 3, 6, 12}.
lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2))*((19+6*sqrt(2))/17)/((9+4*sqrt(2))/7)^3 for n mod 15 = {4, 8, 10, 14}.
lim_{n -> infinity} a(n)/a(n-1) = ((9+4*sqrt(2))/7)^4/((3+2*sqrt(2))*((19+6*sqrt(2))/17)^2) for n mod 15 = 9.

			124^2+(124+833)^2 = 15376+915849 = 931225 = 965^2.

Cf. A156835, A076296, A118120, A118554, A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7), A156163 (decimal expansion of (19+6*sqrt(2))/17).

{forstep(n=0, 50000, [3, 1], if(issquare(2*n^2+1666*n+693889), print1(n, ",")))}

a(n) = 6*a(n-15)-a(n-30)+1666 for n > 30; a(1) = 0, a(2) = 124, a(3) = 168, a(4) = 187, a(5) = 343, a(6) = 399, a(7) = 595, a(8) = 624, a(9) = 915, a(10) = 952, a(11) = 1260, a(12) = 1372, a(13) = 1768, a(14) = 1827, a(15) = 1975, a(16) = 2499, a(17) = 3135, a(18) = 3367, a(19) = 3468, a(20) = 4312, a(21) = 4620, a(22) = 5712, a(23) = 5875, a(24) = 7524, a(25) = 7735, a(26) = 9499, a(27) = 10143, a(28) = 12427, a(29) = 12768, a(30) = 13624.

G.f.: x*(124+44*x+19*x^2+156*x^3+56*x^4+196*x^5+29*x^6+291*x^7+37*x^8+308*x^9+112*x^10+396*x^11+59*x^12+148*x^13+524*x^14-108*x^15-32*x^16-13*x^17-92*x^18-28*x^19-84*x^20-11*x^21-97*x^22-11*x^23-84*x^24-28*x^25-92*x^26-13*x^27-32*x^28-108*x^29)/((1-x)*(1-6*x^15+x^30)).

Edited by Klaus Brockhaus, Feb 16 2009

A201916 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+2737)^2 = y^2.

Original entry on oeis.org

0, 75, 203, 323, 552, 708, 1020, 1127, 1311, 1428, 1608, 1820, 1955, 2336, 2675, 3128, 3311, 3627, 3927, 4140, 4508, 4743, 5535, 6003, 6800, 7280, 7848, 8211, 8588, 9240, 9860, 11063, 11895, 13583, 14168, 15180, 15827, 16827, 18011, 18768, 20915, 22836

Offset: 1

T. D. Noe, Feb 09 2012

Note that 2737 = 7 * 17 * 23, the product of the first three distinct primes in A058529 (and A001132) and hence the smallest such number. This sequence satisfies a linear difference equation of order 55 whose 55 initial terms can be found by running the Mathematica program.
There are many sequences like this one. What determines the order of the linear difference equation? All primes p have order 7. For those p, it appears that p^2 has order 11, p^3 order 15, and p^i order 3+4*i. It appears that for semiprimes p*q (with p > q), the order is 19. What is the next term of the sequence beginning 3, 7, 19, 55, 163? This could be sequence A052919, which is 1 + 2*3^f, where f is the number of primes.
The crossref list is thought to be complete up to Feb 14 2012.

Colin Barker, Table of n, a(n) for n = 1..1000

Cf. A001652 (1), A076296 (7), A118120 (17), A118337 (23), A118674 (31).
Cf. A129288 (41), A118675 (47), A118554 (49), A118673 (71), A129289 (73).
Cf. A118676 (79), A129298 (89), A129836 (97), A157119 (103), A161478 (113).
Cf. A129837 (119), A129992 (127), A129544 (137), A161482 (151).
Cf. A206426 (161), A130608 (167), A161486 (191), A185394 (193).
Cf. A129993 (199), A198294 (217), A130609 (223), A129625 (233).
Cf. A204765 (239), A129991 (241), A207058 (263), A129626 (281).
Cf. A205644 (287), A207059 (289), A129640 (313), A205672 (329).
Cf. A129999 (337), A118611 (343), A130610 (359), A207060 (401).
Cf. A129641 (409), A207061 (433), A130645 (439), A130004 (449).
Cf. A129642 (457), A129725 (521), A101152 (569), A130005 (577).
Cf. A207075 (479), A207076 (487), A207077 (497), A207078 (511).
Cf. A111258 (601), A115135 (617), A130013 (647), A130646 (727).
Cf. A122694 (761), A123654 (809), A129010 (833), A130647 (839).
Cf. A129857 (857), A130014 (881), A129974 (937), A129975 (953).
Cf. A130017 (967), A118630 (2401), A118576 (16807).

d = 2737; terms = 100; t = Select[Range[0, 55000], IntegerQ[Sqrt[#^2 + (#+d)^2]] &]; Do[AppendTo[t, t[[-1]] + 6*t[[-27]] - 6*t[[-28]] - t[[-54]] + t[[-55]]], {terms-55}]; t

a(n) = a(n-1) + 6*a(n-27) - 6*a(n-28) - a(n-54) + a(n-55), where the 55 initial terms can be computed using the Mathematica program.

G.f.: x^2*(73*x^53 +116*x^52 +100*x^51 +171*x^50 +104*x^49 +184*x^48 +57*x^47 +92*x^46 +55*x^45 +80*x^44 +88*x^43 +53*x^42 +139*x^41 +113*x^40 +139*x^39 +53*x^38 +88*x^37 +80*x^36 +55*x^35 +92*x^34 +57*x^33 +184*x^32 +104*x^31 +171*x^30 +100*x^29 +116*x^28 +73*x^27 -363*x^26 -568*x^25 -480*x^24 -797*x^23 -468*x^22 -792*x^21 -235*x^20 -368*x^19 -213*x^18 -300*x^17 -316*x^16 -183*x^15 -453*x^14 -339*x^13 -381*x^12 -135*x^11 -212*x^10 -180*x^9 -117*x^8 -184*x^7 -107*x^6 -312*x^5 -156*x^4 -229*x^3 -120*x^2 -128*x -75) / ((x -1)*(x^54 -6*x^27 +1)). - Colin Barker, May 18 2015

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A118673 Positive solutions x to the equation x^2 + (x+71)^2 = y^2.

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A076293 Numbers k where the root mean square (RMS) of k and 7 is an integer, i.e., sqrt((k^2 + 7^2)/2) is an integer.

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A076294 Consider all Pythagorean triples (X,X+7,Z); sequence gives Z values.

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A076295 Consider all Pythagorean triples (Y-7,Y,Z); sequence gives Y values.

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A129010 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+833)^2 = y^2.

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A201916 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+2737)^2 = y^2.

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