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.

User: Carmine Suriano

Carmine Suriano's wiki page.

Carmine Suriano has authored 156 sequences. Here are the ten most recent ones:

A355554 Sexagesimal expansion of 180/Pi.

Original entry on oeis.org

57, 17, 44, 48, 22, 29, 22, 22, 7, 32, 46, 14, 58, 15, 20, 17, 32, 7, 4, 43, 35, 36, 12, 35, 9, 17, 4, 12, 9, 40, 27, 27, 26, 48, 25, 12, 52, 48, 52, 18, 21, 42, 13, 53, 32, 25, 44, 46, 54, 25, 56, 34, 21, 51, 6, 35, 33, 34, 49, 6, 43, 10, 36, 31, 50, 20, 31
Offset: 1

Author

Carmine Suriano, Jan 17 2023

Keywords

Comments

1 radian = 180/Pi degrees (57 degrees 17' 44.806247...").

Examples

			a(3) = floor((((180/Pi-57)*60)-17)*60) = 44.
		

Crossrefs

Programs

  • Mathematica
    RealDigits[180/Pi, 60, 100][[1]] (* Amiram Eldar, Feb 03 2023 *)

A272376 Twin primes both of which are the sum of three positive cubes.

Original entry on oeis.org

2267, 2269, 3527, 3529, 10331, 10333, 14867, 14869, 17207, 17209, 18521, 18523, 18917, 18919, 20231, 20233, 20357, 20359, 25577, 25579, 27791, 27793, 28547, 28549, 31247, 31249, 35279, 35281, 36899, 36901, 40697, 40699, 44279, 44281, 48779, 48781, 51479, 51481
Offset: 1

Author

Carmine Suriano, Apr 28 2016

Keywords

Examples

			3527 and 3529 are terms since 3527=3^3+5^3+15^3 and 3529=1^3+11^3+13^3.
		

Crossrefs

Programs

  • Mathematica
    cu[n_] := {}!=Quiet@ IntegerPartitions[n,{3},Range[n^(1/3)]^3, 1]; Flatten@ Rest@ Reap@ Do[If[ PrimeQ[p+2] && cu[p] && cu[p+2], Sow[{p, p+2}]], {p, Prime@ Range@ 10000}] (* Giovanni Resta, Apr 28 2016 *)
  • PARI
    list(lim)=my(v=List(), k, t); lim\=1; for(x=1, sqrtnint(lim-2, 3), for(y=1, min(sqrtnint(lim-x^3-1, 3), x), k=x^3+y^3; for(z=1, min(sqrtnint(lim-k, 3), y), if(isprime(t=k+z^3), listput(v, t))))); v=Set(v); for(i=2,#v-1,if(v[i]!=v[i-1]+2 && v[i]!=v[i+1]-2, v[i]=0)); v=Set(v); v[3..#v] \\ Charles R Greathouse IV, Apr 29 2016

A254226 Numbers n such that the n-th decimal digit of e = 1 + the n-th decimal digit of Pi.

Original entry on oeis.org

25, 28, 31, 35, 40, 61, 73, 107, 114, 130, 135, 138, 148, 151, 152, 170, 188, 192, 195, 202, 209, 218, 222, 231, 234, 238, 263, 265, 290, 297, 315, 378, 423, 432, 436, 444, 448, 457, 470, 481, 485, 489, 490, 526, 537, 546
Offset: 1

Author

Carmine Suriano, Jan 27 2015

Keywords

Examples

			31 is in the sequence since the 31st decimal digit of e is 6 and the 31st decimal digit of Pi is 5.
		

Crossrefs

Programs

  • Mathematica
    nn = 550; rdpi = RealDigits[Pi, 10, nn][[1]]; rde = RealDigits[E, 10, nn][[1]]; Select[ Range[2, nn], 1 + rdpi[[#]] == rde[[#]] &] - 1 (* Robert G. Wilson v, Feb 08 2015 *)

A254227 Numbers such that the n-th decimal digit of Pi is one larger than the n-th decimal digit of e.

Original entry on oeis.org

5, 6, 10, 19, 26, 56, 78, 82, 96, 111, 126, 127, 160, 166, 172, 174, 216, 220, 227, 240, 242, 246, 249, 250, 272, 276, 286, 299, 314, 332, 339, 346, 351, 352, 358, 362, 365, 372, 382, 396, 402, 405, 425, 430, 442, 468, 474, 486, 487, 502
Offset: 1

Author

Carmine Suriano, Jan 27 2015

Keywords

Examples

			a(2) = 6 since the 6th decimal digit of Pi is 2 and the 6th decimal digit of e is 1.
		

Crossrefs

Programs

  • Mathematica
    max = 510; piDigits = RealDigits[Pi, 10, max][[1]]; eDigits = RealDigits[E, 10, max][[1]]; Select[Range[2, max], piDigits[[#]] == 1 + eDigits[[#]] &] - 1 (* Robert G. Wilson v, Feb 08 2015 *)

A254836 Numbers n expressible as a product of 4 factors in two different ways, n = a*b*c*d = x*y*w*z, with a+b+c+d = x+y+w+z.

Original entry on oeis.org

36, 40, 48, 72, 80, 90, 96, 108, 120, 126, 144, 160, 168, 176, 180, 192, 200, 216, 225, 234, 240, 252, 270, 280, 288, 297, 300, 320, 324, 336, 352, 360, 378, 384, 396, 400, 405, 408, 420, 432, 440, 448, 450, 456, 468, 480
Offset: 1

Author

Carmine Suriano, Feb 08 2015

Keywords

Comments

The first few terms have a+b+c+d = w+x+y+z equal to 14, 15, 12, 15, 16, 17, 15, ..., . - Robert G. Wilson v, Feb 09 2015

Examples

			40 is in the list since 40 = 1*1*5*8 = 1*2*2*10 and 1+1+5+8 = 15 = 1+2+2+10.
		

Crossrefs

Cf. A060292.

Programs

  • Mathematica
    fQ[n_] := If[ PrimeOmega@ n > 3, Block[{k = 1}, While[k < n && Length@ Select[ IntegerPartitions[k, {4}, Divisors@ n], Times @@ # == n &] < 2, k++]; If[k < 2n, True]]]; k = 1; lst = {}; While[k < 500, If[ fQ@ k, AppendTo[lst, k]]; k++]; lst (* Robert G. Wilson v, Feb 09 2015 *)

A253408 Values of difference z-y that solve equation x^2 + y^2 = z^2 + 2.

Original entry on oeis.org

-1, 1, 7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97, 103, 113, 119, 127, 137, 151, 161, 167, 191, 193, 199, 217, 223, 233, 239, 241, 257, 263, 271, 281, 287, 289, 311, 313, 329, 337, 343, 353, 359, 367, 383, 391, 401, 409, 431, 433, 439, 449, 457, 463, 479, 487
Offset: 1

Author

Carmine Suriano, Jan 05 2015

Keywords

Comments

All values are congruent to +-1 (mod 8).

Examples

			a(3)=7 because 25^2 + 41^2 = 48^2 + 2 and 48 - 41 = 7.
		

Crossrefs

Cf. A058529.

Formula

a(n)^2 mod 8 = 1.
a(n) = A058529(n-1), n>1.

A246666 Numbers n such that n^3 + (n+1)^3 + (n+3)^3 is prime.

Original entry on oeis.org

1, 3, 5, 9, 31, 43, 45, 51, 71, 89, 135, 141, 145, 149, 159, 163, 169, 183, 185, 225, 233, 241, 255, 261, 271, 281, 283, 285, 299, 309, 311, 313, 355, 395, 401, 411, 415, 423, 429, 435, 449, 453, 485, 491, 541, 551, 561, 579, 583, 589, 603, 621, 625, 635, 681
Offset: 1

Author

Carmine Suriano, Sep 01 2014

Keywords

Comments

All terms are odd. - Jon Perry, Sep 11 2014

Examples

			a(5)=31 since 31^3+(31+1)^3+(31+3)^3=101863 is prime.
		

Programs

  • Maple
    isA246666 := proc (n) return isprime(n^3+(n+1)^3+(n+3)^3) end proc; seq(`if`(isA246666(2*n-1), 2*n-1, NULL), n = 1 .. 400); # Nathaniel Johnston, Sep 09 2014
  • PARI
    for(n=0,10^3,if(isprime(n^3+(n+1)^3+(n+3)^3),print1(n,", "))); \\ Joerg Arndt, Sep 09 2014
    
  • Python
    from sympy import isprime
    A246666_list = [n for n in range(1,10**5) if isprime(3*n*(n*(n+4)+10)+28)]
    # Chai Wah Wu, Sep 09 2014

A242538 Squares that are sum of digits of factorials.

Original entry on oeis.org

1, 9, 36, 81, 144, 225, 324, 441, 729, 1089, 1296, 2025, 2304, 2601, 2916, 3249, 3600, 3969, 4356, 4761, 6561, 7056, 7569, 8100, 8649, 9216, 9801, 10404, 11025, 11664, 12996, 13689, 15129, 16641, 17424, 20736, 22500, 23409, 26244, 29241
Offset: 1

Author

Carmine Suriano, May 17 2014

Keywords

Comments

Intersection of A000290 and A004152.

Examples

			a(5)=144 for 33!=8683317618811886495518194401280000000 whose sum of digits is 144=12^2. a(5) is also originated from 34! and 35!.
		

Crossrefs

Cf. A066235.

Programs

  • Mathematica
    Union[Select[Total[IntegerDigits[#]]&/@(Range[2500]!),IntegerQ[Sqrt[#]]&]] (* Harvey P. Dale, Feb 20 2015 *)
  • PARI
    lista(nn) = {v = vector(nn, n, sumdigits(n!)); Set(select(x->issquare(x), v));} \\ Michel Marcus, May 18 2014

A242130 Sum of squares of three numbers x, y, z such that 3*floor(sum of squares/3) = x+y+z.

Original entry on oeis.org

3, 12, 14, 18, 27, 29, 33, 35, 41, 48, 50, 54, 56, 62, 66, 72, 74, 75, 77, 81, 83, 89, 93, 99, 101, 107, 108, 110, 114, 116, 122, 126, 132, 134, 140, 146, 147, 149, 153, 155, 161, 165, 171, 173, 179, 185, 189, 192, 194, 198, 200
Offset: 1

Author

Carmine Suriano, May 05 2014

Keywords

Comments

Three times the quadratic average of x,y,z in most cases is greater than their sum.

Examples

			a(7)=33 for 33=1^2+4^2+4^2; floor(sqrt(33/3))=3; 3*3=9=1+4+4.
		

Crossrefs

Cf. A000408.

A239064 Positive integers whose square can be expressed as k*x^2+(k+1)*x+(k+2) with k and x positive integers.

Original entry on oeis.org

3, 5, 6, 9, 10, 12, 15, 16, 18, 19, 21, 23, 24, 26, 27, 30, 33, 36, 37, 39, 40, 41, 42, 44, 45, 47, 48, 51, 52, 54, 57, 58, 60, 61, 63, 64, 65, 66, 67, 68, 69, 72, 75, 78, 79, 81, 82, 83, 84, 86, 87, 89, 90, 92, 93, 96, 99, 100, 102, 103, 105, 107, 108, 110, 111, 114, 117, 120, 121, 123, 124, 126, 128, 129, 130, 131, 132, 134, 135, 138, 141, 142, 144, 145, 147, 149, 150, 152, 153, 156, 159, 162, 163, 165, 166, 168, 170, 171, 173, 174, 176, 177, 180, 183, 184, 186, 187, 189, 191, 192, 194, 195, 196, 198, 201, 204
Offset: 1

Author

Carmine Suriano, Mar 28 2014

Keywords

Comments

Some can be expressed in more than one way. E.g., a(46) = 81 for x = 1, k = 26 and for x = 2 and k = 11.
x=1 for k=3*m^2-1 that is A080663, with a = 3*m.
From Wolfdieter Lang, Apr 26 2014: (Start)
The solution of this problem can be found as follows. Consider k*x^2 + (k+1)*x + (k+2) - a^2 = 0. Solve for x (positive) as function of k: x = (-(k+1) + sqrt((2*a)^2*k - c(k)))/(2*k), where c(k) = 3*k(k+2) - 1 is the sequence [8, 23, 44, 71, 104, 143, 188, 239, 296, 359,...]. A necessary condition for a solution is (2*a)^2*k - c(k) = b^2, or k*X^2 - Y^2 = c(k), with X = 2*a and Y = b. This is a binary indefinite quadratic form with discriminant D = (2^2)*k > 0. If k is a square then there is no solution with even X because (K*X)^2 - Y^2 = c(K^2) has only the solution with K=1, X=3, Y=1. For k not a square there are either no solutions for given k (e.g., k = 5, 6, 10, ...) or countable infinitely many ones of the representation k*X^2 - Y^2 = c(k). From the solutions one has to pick first the ones with even X (a = X/2) and then to test whether 2*k divides Y - (k+1). The solution for x is then x = (Y - (k+1))/(2*k).
For representations of integers by indefinite binary quadratic forms see, e.g., the D. A. Buell reference, and also a W. Lang link (with further references) with an on-line program.
(End)

Examples

			a(10)=19 from 19^2=361=51*2^2+(51+1)*2+(51+2).
From _Wolfdieter Lang_, Apr 21 2014: (Start)
k = 2: There are two proper solutions of 2*X^2 - Y^2 = 23, namely [4, 3], [6, 7]. Both generate infinitely many new solutions, all with even X, namely [4, 3], [18, 25], [104, 147], [606, 857],[3532, 4995], [20586, 29113], ... and [6, 7], [32, 45], [186, 263], [1084, 1533],[6318, 8935], [36824, 52077], ... . Only every other solution has 2*k = 4 dividing Y - (2+1) = Y-3, giving the positive solutions for (a=X/2, b=Y; x), starting with the second proper solution and then alternating between the two sets of solutions (3, 7; 1), (52, 147; 36), (93, 263; 65), (1766, 4995; 1248), (3159, 8935; 2233), ... . Thus the positive x solutions for k = 2 are 1, 36, 65, 1248, 2233, ..., with a = 3, 52, 93, 1766, 3159, ... .
k = 3: the positive solutions for x are 2, 5, 38, 79, 538, 1109, ..., with a = 5, 10, 67, 138, 933, 1922, ..., coming from the even X solutions of 3*X^2 - Y^2 = 44, [4, 2], [10, 16], [36, 62], [134, 232], [500, 866], [1866, 3232], ... and [6, 8], [20, 34], [74, 128], [276, 478], [1030, 1784], [3844, 6658], ... . Then 2*k = 6 has to divide Y - 4, leaving every other of these solutions with (a = X/2, b=Y; x) given by (5, 16; 2), (10, 34; 5), (67, 232; 38), (138, 478; 79), (933, 3232; 538), (1922, 6658; 1109), ... .
k = 5: there are no solutions of 5*X^2 - Y2 = 104.
k = 6: there are no solutions of 6*X^2 - Y2 = 143.
(End)
		

References

  • D. A. Buell, Binary Quadratic Forms. Springer-Verlag, NY, 1989.

Programs

  • BASIC
    for k=1 to z
      for x=1 to z
        a=k*x*x+(k+1)*x+(k+2)
        if sqr(a)-int(sqr(a))=0 then
          j=j+1
          a_n(j)=sqr(a)
        endif
       next x
    next k

Formula

Increasingly ordered set {a from the positive integers: a^2 = k*x^2 + (k+1)*x + (k+2), with k and x positive integers}. - Wolfdieter Lang, Apr 21 2014

Extensions

Edited, name specified, comment changed, reference and link added by Wolfdieter Lang, Apr 21 2014