Mohamed Bouhamida - cp's OEIS frontend

A331265 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 31^2)^2 = y^2.

0, 279, 656, 1139, 1860, 2883, 4340, 6419, 9156, 13299, 19220, 27683, 39780, 55719, 79856, 114359, 163680, 234183, 327080, 467759, 668856, 956319, 1367240, 1908683, 2728620, 3900699, 5576156, 7971179, 11126940, 15905883, 22737260, 32502539, 46461756, 64854879, 92708600, 132524783

Offset: 1

Mohamed Bouhamida, Feb 12 2020

For the generic case x^2 + (x + p^2)^2 = y^2 with p = 2*m^2 - 1 a (prime) number in A066436, m >= 4 (means p >= 31), the first five consecutive solutions are (0, p^2), (4*m^3+2*m^2-2*m-1, 4*m^4+4*m^3-2*m-1), (8*m^3+8*m^2+4*m, 4*m^4+8*m^3+12*m^2+4*m+1), (12*m^4-40*m^3+44*m^2-20*m+3, 20*m^4-56*m^3+60*m^2-28*m+5), (12*m^4-20*m^3+2*m^2+10*m-4, 20*m^4-28*m^3+14*m-5) and the other solutions are defined by (X(n), Y(n)) = (3*X(n-5) + 2*Y(n-5) + p^2, 4*X(n-5) + 3*Y(n-5) + 2*p^2).

X(n) = 6*X(n-5) - X(n-10) + 2*p^2, and Y(n) = 6*Y(n-5) - Y(n-10) (can be easily proved using X(n) = 3*X(n-5) + 2*Y(n-5) + p^2, and Y(n) = 4*X(n-5) + 3*Y(n-5) + 2*p^2).

			For p=31 (m=4) the first five (5) consecutive solutions are (0, 961), (279, 1271), (656, 1745), (1139, 2389), (1860, 3379).

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

Cf. A066436 (Primes of the form 2*m^2 - 1).

Solutions x to x^2+(x+p^2)^2=y^2: A118554 (p=7), A207059 (p=17), A309998 (p=23), this sequence (p=31), A332000 (p=47).

I:=[0, 279, 656, 1139, 1860, 2883, 4340, 6419, 9156, 13299]; [n le 10 select I[n] else 6*Self(n-5) - Self(n-10)+1922: n in [1..100]];

LinearRecurrence[{1, 0, 0, 0, 6, -6, 0, 0, 0, -1, 1}, {0, 279, 656, 1139, 1860, 2883, 4340, 6419, 9156, 13299, 19220}, 36] (* Jean-François Alcover, Feb 12 2020 *)

concat(0, Vec(x^2*(279 + 377*x + 483*x^2 + 721*x^3 + 1023*x^4 - 217*x^5 - 183*x^6 - 161*x^7 - 183*x^8 - 217*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)) + O(x^30))) \\ Colin Barker, Feb 12 2020

a(n) = 6*a(n-5) - a(n-10) + 1922 for n >= 11; a(1)=0, a(2)=279, a(3)=656, a(4)=1139, a(5)=1860, a(6)=2883, a(7)=4340, a(8)=6419, a(9)=9156, a(10)=13299.
From Colin Barker, Feb 12 2020: (Start)
G.f.: x^2*(279 + 377*x + 483*x^2 + 721*x^3 + 1023*x^4 - 217*x^5 - 183*x^6 - 161*x^7 - 183*x^8 - 217*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)).
a(n) = a(n-1) + 6*a(n-5) - 6*a(n-6) - a(n-10) + a(n-11) for n>11.
(End)

A332000 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 47^2)^2 = y^2.

Original entry on oeis.org

0, 752, 1820, 2231, 3995, 6627, 10575, 16511, 18840, 28952, 44180, 67116, 101664, 115227, 174135, 262871, 396539, 597891, 676940, 1020276, 1537464, 2316536, 3490100, 3950831, 5951939, 8966331, 13507095, 20347127, 23032464, 34695776, 52264940, 78730452, 118597080, 134248371

Offset: 1

Mohamed Bouhamida, Feb 04 2020

For the generic case x^2 + (x + p^2)^2 = y^2 with p = m^2 - 2 a (prime) number in A028871, m>=7 (means p>=47), the first five consecutive solutions are: (0; p^2), (2*m^3+2*m^2-4*m-4; m^4+2*m^3-4*m-4), (4*m^3+8*m^2+8*m; m^4+4*m^3+12*m^2+8*m+4), (3*m^4-20*m^3+44*m^2-40*m+12; 5*m^4-28*m^3+60*m^2-56*m+20), (3*m^4-10*m^3+2*m^2+20*m-16; 5*m^4-14*m^3+28*m-20) and the other solutions are defined by: (X(n); Y(n))= (3*X(n-5)+2*Y(n-5)+p^2; 4*X(n-5)+3*Y(n-5)+2*p^2).

X(n) = 6*X(n-5) - X(n-10) + 2*p^2, and Y(n) = 6*Y(n-5) - Y(n-10) (can be easily proved using X(n) = 3*X(n-5) + 2*Y(n-5) + p^2, and Y(n) = 4*X(n-5) + 3*Y(n-5) + 2*p^2).

			For p=47 (m=7) the first five (5) consecutive solutions are (0, 2209), (752, 3055), (1820, 4421), (2231, 4969), (3995, 7379).

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

Cf. A028871 (Primes of the form m^2 - 2).

Solutions x to x^2+(x+p^2)^2=y^2: A118554 (p=7), A207059 (p=17), A309998 (p=23), A331265 (p=31), this sequence (p=47).

I:=[0, 752, 1820, 2231, 3995, 6627, 10575, 16511, 18840, 28952]; [n le 10 select I[n] else 6*Self(n-5) - Self(n-10)+4418: n in [1..100]];

LinearRecurrence[{1, 0, 0, 0, 6, -6, 0, 0, 0, -1, 1}, {0, 752, 1820, 2231, 3995, 6627, 10575, 16511, 18840, 28952, 44180}, 40] (* Jean-François Alcover, Feb 08 2020 *)

concat(0, Vec(x^2*(752 + 1068*x + 411*x^2 + 1764*x^3 + 2632*x^4 - 564*x^5 - 472*x^6 - 137*x^7 - 472*x^8 - 564*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)) + O(x^40))) \\ Colin Barker, Feb 04 2020

a(n) = 6*a(n-5) - a(n-10) + 4418 for n >= 11; a(1)=0, a(2)=752, a(3)=1820, a(4)=2231, a(5)=3995, a(6)=6627, a(7)=10575, a(8)=16511, a(9)=18840, a(10)=28952.
From Colin Barker, Feb 04 2020: (Start)
G.f.: x^2*(752 + 1068*x + 411*x^2 + 1764*x^3 + 2632*x^4 - 564*x^5 - 472*x^6 - 137*x^7 - 472*x^8 - 564*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)).
a(n) = a(n-1) + 6*a(n-5) - 6*a(n-6) - a(n-10) + a(n-11) for n>11.
(End)

A309998 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 529)^2 = y^2.

Original entry on oeis.org

0, 276, 287, 740, 759, 1587, 3059, 3120, 5687, 5796, 10580, 19136, 19491, 34440, 35075, 62951, 112815, 114884, 202011, 205712, 368184, 658812, 670871, 1178684, 1200255, 2147211, 3841115, 3911400, 6871151, 6996876, 12516140, 22388936, 22798587, 40049280, 40782059, 72950687

Offset: 1

Mohamed Bouhamida, Aug 26 2019

For the generic case x^2 + (x + p^2)^2 = y^2 with p = m^2 - 2 a prime number in A028871, m>=5, (0; p^2) and (2*m^3 + 2*m^2 - 4*m - 4; m^4 + 2*m^3 - 4*m - 4) are solutions.

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

Cf. A207059.

Rest@ CoefficientList[Series[x^2*(276 + 11 x + 453 x^2 + 19 x^3 + 828 x^4 - 184 x^5 - 5 x^6 - 151 x^7 - 5 x^8 - 184 x^9)/((1 - x) (1 - 6 x^5 + x^10)), {x, 0, 36}], x] (* Michael De Vlieger, Sep 29 2019 *)

concat(0, Vec(x^2*(276 + 11*x + 453*x^2 + 19*x^3 + 828*x^4 - 184*x^5 - 5*x^6 - 151*x^7 - 5*x^8 - 184*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)) + O(x^40))) \\ Colin Barker, Aug 27 2019

a(n) = 6*a(n-5) - a(n-10) + 1058 with a(0) = 0, a(1) = 276, a(2) = 287, a(3) = 740, a(4) = 759, a(5) = 1587, a(6) = 3059, a(7) = 3120, a(8) = 5687, a(9) = 5796.
From Colin Barker, Aug 27 2019: (Start)
G.f.: x^2*(276 + 11*x + 453*x^2 + 19*x^3 + 828*x^4 - 184*x^5 - 5*x^6 - 151*x^7 - 5*x^8 - 184*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)).
a(n) = a(n-1) + 6*a(n-5) - 6*a(n-6) - a(n-10) + a(n-11) for n>11.
(End)

A164643 Semiprimes pq with pq - 1 divisible by p + q.

Original entry on oeis.org

6, 21, 301, 697, 1333, 1909, 2041, 3901, 24601, 26977, 96361, 130153, 163201, 250321, 275833, 296341, 389593, 486877, 495529, 542413, 808861, 1005421, 1005649, 1055833, 1063141, 1232053, 1284121, 1403221, 1618597, 1787917, 2287933, 2462881, 2488201, 2666437

Offset: 1

Mohamed Bouhamida, Aug 19 2009

nonn

The first three terms are Syl(0)*Syl(1), Syl(1)*Syl(2) and Syl(2)*Syl(3). Syl means Sylvester's sequence, see A000058.
Products of two consecutive numbers p and q in Sylvester's sequence with primes p and q are in the sequence.
Let p and q be consecutive prime Sylvester numbers. Then: pq - 1 = p*(p^2 - p + 1) - 1 = p^3 - p^2 + p - 1 = (p^2 + 1)*(p - 1) = (p + p^2 - p + 1)*(p - 1) = (p + q)*(p - 1) it means that: (pq - 1) is divisible by (p + q). - Mohamed Bouhamida, Aug 21 2009
(p-k)*(q-k) = k^2 + 1 for some integer k, providing a fast way for finding appropriate p,q. - Max Alekseyev, Aug 26 2009

Donovan Johnson, Table of n, a(n) for n = 1..1000

Cf. A001358, A000058.

isA001358 := proc(n) RETURN ( numtheory[bigomega](n) =2 ) ; end:
isA164643 := proc(n) if isA001358(n) then p := op(1,op(1,ifactors(n)[2]) ) ; q := n/p ; if (p*q-1) mod (p+q) =0 then true; else false; fi; else false; fi; end:
for n from 4 to 3000000 do if isA164643(n) then print(n) ; fi; od: # R. J. Mathar, Aug 24 2009

dsQ[n_]:=Module[{prs=Transpose[FactorInteger[n]][[1]]},Divisible[n-1, Total[prs]]]; Select[Select[Range[2000000], PrimeOmega[#] ==2&], dsQ] (* Harvey P. Dale, Jun 15 2011 *)

Extended by R. J. Mathar, Aug 24 2009

More terms from Max Alekseyev, Aug 26 2009

A164698 Semiprimes pq such that pq - 1 divides p^2 + q^2 + 2.

Original entry on oeis.org

6, 21, 26, 51, 1157, 372101, 1288005205276048901

Offset: 1

Mohamed Bouhamida, Aug 22 2009

Semiprimes pq such that pq-1 divides (p+q)^2.
The third to fifth terms are Fib(3)*Fib(7), Fib(7)*Fib(11) and Fib(13)*Fib(17).
Products of two prime Fibonacci numbers F(k) and F(k+4) (see A001605 and A005478) are in the sequence.
6 and 26 are the only even terms. Odd terms contain products of pairs of consecutive terms from the following sequences: A005248, A001541, A033889, A033891. - Max Alekseyev, Aug 27 2009

			The semiprime 6 = 2*3 is in the sequence because 2*3 - 1 = 5 divides 2^2 + 3^2 + 2 = 15.

Cf. A001358, A000045, A140362, A164643.

isA001358 := proc(n) RETURN ( numtheory[bigomega](n) =2 ) ; end:
isA164698 := proc(n) if isA001358(n) then p := op(1,op(1,ifactors(n)[2]) ) ; q := n/p ; if (p^2+q^2+2) mod (p*q-1) = 0 then true; else false; fi; else false; fi; end:
for n from 4 to 3000000 do if isA164698(n) then print(n, ifactors(n)) ; fi; od: # R. J. Mathar, Aug 24 2009

Missing values added by R. J. Mathar, Aug 24 2009

a(7) from Max Alekseyev, Aug 27 2009

A140362 Semiprimes pq that divide the sum of the squares of their divisors, 1+p^2+q^2+(pq)^2.

Original entry on oeis.org

10, 65, 20737

Offset: 1

Mohamed Bouhamida, Jul 22 2008, Jul 27 2008

6 is the smallest integer n which is the product of two distinct primes and which divides the sum of the cubes of the divisors of n. Are there other numbers with this property?
Using Pell equations and a Fibonacci identity, Max Alekseyev and I have shown that all terms are the product of prime Fibonacci numbers whose indices are twin primes. The first three terms are Fib(3)*Fib(5), Fib(5)*Fib(7) and Fib(11)*Fib(13). The other two known terms are Fib(431)*Fib(433) and Fib(569)*Fib(571), huge numbers that are in the b-file. The sequence probably has no additional terms. - T. D. Noe, Jul 27 2008
Let a, b, c and d be consecutive odd-indexed Fibonacci numbers. Then it can be proved that 1 + b^2 + c^2 + (bc)^2 = abcd, which shows that bc divides 1 + b^2 + c^2 + (bc)^2. Hence if b and c are prime, then bc is in this sequence. - T. D. Noe, Jul 27 2008
Empirical search suggests that A067558(a(n))/A032741(a(n)) = a(n). A032741(a(n)) = 3 for all n by definition of semiprime. A067558(a(n)) must also then be divisible by 3. a(n) can be called the n-th "perfect mean square aliquot number". - William Krier, Dec 16 2024

			10 divides (1^2 + 2^2 + 5^2) giving 3 - the number of proper divisors of semiprime 10.
65 divides (1^2 + 5^2 + 13^2) giving 3 - the number of proper divisors of semiprime 65.
20737 divides (1^2 + 89^2 + 233^2) giving 3 - the number of proper divisors of semiprime 20737.

T. D. Noe, Table of n, a(n) for n=1..5
T. Cai, D. Chen, and Y. Zhang, Perfect numbers and Fibonacci primes, arXiv:1310.0898 [math.NT], 2013-2014.
T. Cai, D. Chen, and Y. Zhang, Perfect numbers and Fibonacci primes (II), arXiv:1406.5684 [math.NT], 2014 (see case m=1 in Table 1).

Cf. A000045, A001605, A046762.

isok(n) = sigma(n, 2) - n^2 == 3*n; \\ Michel Marcus, Jun 24 2014

A135432 a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) - a(n-5) with a(0)=0, a(1)=1, a(2)=2, a(3)=3 and a(4)=4.

Original entry on oeis.org

0, 1, 2, 3, 4, 10, 18, 33, 62, 119, 222, 418, 788, 1485, 2794, 5263, 9912, 18666, 35150, 66197, 124662, 234763, 442106, 832578, 1567912, 2952697, 5560530, 10471611, 19720172, 37137098, 69936714, 131705065, 248027438

Offset: 0

Mohamed Bouhamida, Dec 13 2007

nonn

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

LinearRecurrence[{1, 1, 1, 1, -1}, {0, 1, 2, 3, 4}, 50] (* G. C. Greubel, Oct 14 2016 *)

G.f.: -x*(x-1)*(2*x^2+2*x+1)/((x+1)*(x^4-2*x^3+x^2-2*x+1)). - Maksym Voznyy, Aug 12 2009

A135431 a(n) = a(n-1) + a(n-2) + a(n-3) - a(n-4) with a(0)=0, a(1)=1, a(2)=2 and a(3)=3.

Original entry on oeis.org

0, 1, 2, 3, 6, 10, 17, 30, 51, 88, 152, 261, 450, 775, 1334, 2298, 3957, 6814, 11735, 20208, 34800, 59929, 103202, 177723, 306054, 527050, 907625, 1563006, 2691627, 4635208, 7982216, 13746045, 23671842, 40764895, 70200566, 120891258, 208184877, 358511806

Offset: 0

Mohamed Bouhamida, Dec 13 2007

nonn

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

LinearRecurrence[{1, 1, 1, -1}, {0, 1, 2, 3}, 40] (* Vladimir Joseph Stephan Orlovsky, Feb 03 2012 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,d+c+b-a}; NestList[nxt,{0,1,2,3},40][[All,1]] (* Harvey P. Dale, Feb 28 2021 *)

From R. J. Mathar, Mar 31 2008: (Start)
O.g.f.: x*(1+x)/(1-x-x^2-x^3+x^4).
a(n) = A116732(n+1) + A116732(n+2). (End)

A133360 Number of surjections from an n-element set to a nine-element set.

Original entry on oeis.org

362880, 16329600, 419126400, 8083152000, 130456085760, 1863435974400, 24359586451200, 297846188640000, 3457819037312640, 38528927611574400, 415357755774998400, 4358654246117808000, 44733116259693227520

Offset: 9

Mohamed Bouhamida, Dec 21 2007

Index entries for linear recurrences with constant coefficients, signature (45,-870,9450,-63273,269325,-723680,1172700,-1026576,362880).

Column k=9 of A131689.

Cf. A000918, A000919, A000920, A001117, A001118, A049447, A135456.

a(n) = Sum_{k=1..9} (-1)^(9-k)*binomial(9,k)*k^n.
a(n) = A049447(n) * 9!. - Max Alekseyev, Nov 12 2009
G.f.: -362880*x^9/((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)*(8*x-1)*(9*x-1)). - Colin Barker, Oct 25 2012
E.g.f.: (exp(x) - 1)^9. - Ilya Gutkovskiy, Jun 19 2018

More terms from Max Alekseyev, Nov 12 2009

A133132 Number of surjections from an n-element set to a ten-element set.

Original entry on oeis.org

3628800, 199584000, 6187104000, 142702560000, 2731586457600, 45950224320000, 703098107712000, 10009442963520000, 134672620008326400, 1732015476199008000, 21473732319740064000, 258323865658578720000

Offset: 10

Mohamed Bouhamida, Dec 16 2007

Vincenzo Librandi, Table of n, a(n) for n = 10..1000
Index entries for linear recurrences with constant coefficients, signature (55,-1320,18150,-157773,902055,-3416930,8409500,-12753576,10628640,-3628800).

Column k=10 of A131689.

Cf. A000918, A000919, A000920, A001117, A001118, A049435, A135456.

[10^n-10*9^n+45*8^n-120*7^n+210*6^n-252*5^n+210*4^n-120*3^n+45*2^n-10: n in [10..30]]; // Vincenzo Librandi, Apr 11 2012

With[{nn=30},Drop[CoefficientList[Series[(Exp[x]-1)^10,{x,0,nn}],x] Range[0,nn]!,10]] (* Harvey P. Dale, Sep 01 2016 *)

sum(k=1,10,(-1)^(10-k)*binomial(10,k)*k^n)

a(n) = 10^n-10*9^n+45*8^n-120*7^n+210*6^n-252*5^n+210*4^n-120*3^n+45*2^n-10.
a(n) = A049435(n) * 10!. - Max Alekseyev, Nov 13 2009
G.f.: 3628800*x^10/((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)*(8*x-1)*(9*x-1)*(10*x-1)). - Colin Barker, Oct 25 2012
E.g.f.: (exp(x)-1)^10. - Alois P. Heinz, May 17 2016

More terms from Max Alekseyev, Nov 13 2009

Formula corrected by Charles R Greathouse IV, Mar 07 2010

cp's OEIS Frontend

User: Mohamed Bouhamida

A331265 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 31^2)^2 = y^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A332000 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 47^2)^2 = y^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A309998 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 529)^2 = y^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A164643 Semiprimes pq with pq - 1 divisible by p + q.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A164698 Semiprimes pq such that pq - 1 divides p^2 + q^2 + 2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Extensions

A140362 Semiprimes pq that divide the sum of the squares of their divisors, 1+p^2+q^2+(pq)^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A135432 a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) - a(n-5) with a(0)=0, a(1)=1, a(2)=2, a(3)=3 and a(4)=4.

Views

Author

Keywords

Links

Programs

Mathematica

Formula

A135431 a(n) = a(n-1) + a(n-2) + a(n-3) - a(n-4) with a(0)=0, a(1)=1, a(2)=2 and a(3)=3.

Views