A008598 -id:A008598 - cp's OEIS frontend

A334566 Number of solutions of the Diophantine equation z^2 - y^2 - x^2 = n > 0 when the positive integers, x, y and z, are consecutive terms of an arithmetic progression.

0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 3, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 1, 2, 0, 0, 3, 2, 0, 0, 2, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 0, 0, 3, 0, 0, 0, 1, 3, 0, 0, 4, 2, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 3, 1, 0, 0, 1, 1, 0, 0, 1, 3, 0, 0, 2, 0

Offset: 1

Bernard Schott, May 06 2020

nonn

Inspired by the 135th and 136th problems of Project Euler (see links).
If d is the common difference of the arithmetic progression (x, y, z), then the Diophantine equation becomes (y+d)^2 - y^2 - (y-d)^2 = n <==> y^2 - 4dy + n = 0 <==> n = y * (4d-y).
If y is the average term, then y divides n.
Offset is 1 because for n = 0, every (x, y, z)= (3d, 4d, 5d) with d>0 would be solution.

			a(3) = 1 because 4^2 - 3^2 - 2^2 = 3.
a(15) = 3 because 5^2 - 3^2 - 1^2 = 7^2 - 5^2 - 3^2 = 19^2 - 15^2 - 11^2 = 15.
If n = 4q+3, q >= 0 then (3q+2, 4q+3, 5q+4) is a solution.
If n = 16q, q >= 1 then (3q-1, 4q, 5q+1) is a solution.
If n = 16q+4, q >= 0 then (6q+1, 8q+2, 10q+3) is a solution.
If n = 16q+12, q >= 0 then (6q+4, 8q+6, 10q+8) is a solution.

Robert Israel, Table of n, a(n) for n = 1..10000
Project Euler, Problem 135: Same differences
Project Euler, Problem 136: Singleton difference

Cf. A001248, A002145, A004767, A008598, A016813, A016825, A051062.

Cf. A334567 (least value of n such that a(n) = k>0).

f:= proc(n) local r; r:= floor(sqrt(n/3));
nops(select(t -> n/t + t mod 4 = 0 and t > r, numtheory:-divisors(n)))
end proc:
map(f, [$1..100]); # Robert Israel, Jul 31 2020

a[n_] := Length@ Solve[(4 d - x) x == n  && x>0 && x-d>0 && x+d>0, {d, x}, Integers]; Array[a, 90] (* Giovanni Resta, May 06 2020 *)

a(n) = 0 iff n = 4q+1 (A016813), n = 4q+2 (A016825), n = 16q+8 (A051062), q>= 0.
a(n) >= 1 iff n = 4q+3, q >=0 (A004767), n = 16q, q>=1 (A008598), n = 16q+4, q>=0 (A119413), n = 16q+12, q>=0 (A098502).
a(4*q^2) >= 1, for q >= 1, since (q, 2q, 3q) is a solution.
a(p) = 1 for p = 4q+3 prime (A002145).
a(p^2) = 0 for p an odd prime (A065091).

More terms from Giovanni Resta, May 06 2020

A351381 Table read by downward antidiagonals: T(n,k) = n*(k+1)^2.

Original entry on oeis.org

4, 9, 8, 16, 18, 12, 25, 32, 27, 16, 36, 50, 48, 36, 20, 49, 72, 75, 64, 45, 24, 64, 98, 108, 100, 80, 54, 28, 81, 128, 147, 144, 125, 96, 63, 32, 100, 162, 192, 196, 180, 150, 112, 72, 36, 121, 200, 243, 256, 245, 216, 175, 128, 81, 40, 144, 242, 300, 324, 320, 294, 252, 200, 144, 90, 44

Offset: 1

Bernard Schott, Mar 28 2022

When m and k are both positive integers and k | m, with m/k = n, then T(n,k) = S(m,k) = (m+k) + (m-k) + (m*k) + (m/k) = S(n*k,k) = n*(k+1)^2, problem proposed by Yakov Perelman.

All terms are nonsquarefree (A013929).

			Table begins:
  n \ k |   1      2      3      4      5      6      7      8      9     10
  ----------------------------------------------------------------------------
     1  |   4      9     16      25    36     49     64     81    100    121
     2  |   8     18     32      50    72     98    128    162    200    242
     3  |  12     27     48      75   108    147    192    243    300    363
     4  |  16     36     64     100   144    196    256    324    400    484
     5  |  20     45     80     125   180    245    320    405    500    605
     6  |  24     54     96     150   216    294    384    486    600    726
     7  |  28     63    112     175   252    343    448    567    700    847
     8  |  32     72    128     200   288    392    512    648    800    968
     9  |  36     81    144     225   324    441    576    729    900   1089
    10  |  40     90    160     250   360    490    640    810   1000   1210
  ............................................................................
T(3,4) = 75 = 3*(4+1)^2 corresponds to S(3*4,4) = S(12,4) = (12+4) + (12-4) + (12*4) + 12/4 = 75.
S(10,5) = (10+5) + (10-5) + (10*5) + (10/5) = T(10/5,5) = T(2,5) = 72.

I. Perelman, L'Algèbre Récréative, Chapitre IV, Les équations de Diophante, Deux nombres et quatre opérations, Editions en langues étrangères, Moscou, 1959, pp. 101-102.
Ya. I. Perelman, Algebra Can Be Fun, Chapter IV, Diophantine Equations, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131-132.

Ya. I. Perelman, Algebra Can Be Fun, Chapter IV, Diophantine Equations, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131-132.
Wikipedia, Yakov Perelman.

Cf. A013929.
Cf. A000290 \ {0,1} (row 1), A001105 \ {0,2} (row 2), A033428 \ {0,3} (row 3), A016742 \ {0,4} (row 4), A033429 \ {0,5} (row 5), A033581 \ {0,6} (row 6).
Cf. A008586 \ {0} (column 1), A008591 \ {0} (column 2), A008598 \ {0} (column 3), A008607 \ {0} (column 4), A044102 \ {0} (column 5).
Cf. A045991 \ {0} (diagonal).

T[n_, k_] := n*(k + 1)^2; Table[T[k, n - k + 1], {n, 1, 11}, {k, 1, n}] // Flatten (* Amiram Eldar, Mar 29 2022 *)

T(n,k) = n*(k+1)^2.
T(n,n) = (n+1)^3 - (n+1)^2 = A045991(n+1) for n >= 1.
G.f.: x*(1 + y)/((1 - x)^2*(1 - y)^3). - Stefano Spezia, Mar 31 2022

A171272 a(n) = 1 + 4n(1 + 2*n^2)/3.

Original entry on oeis.org

1, 5, 25, 77, 177, 341, 585, 925, 1377, 1957, 2681, 3565, 4625, 5877, 7337, 9021, 10945, 13125, 15577, 18317, 21361, 24725, 28425, 32477, 36897, 41701, 46905, 52525, 58577, 65077, 72041, 79485, 87425, 95877, 104857, 114381, 124465, 135125, 146377, 158237, 170721, 183845

Offset: 0

Paul Curtz, Dec 06 2009

Binomial transform of quasi-finite sequence 1,4,16,16,0,0,... (0 continued).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

[1+4*n*(1+2*n^2)/3: n in [0..40]]; // Vincenzo Librandi, Aug 05 2011

LinearRecurrence[{4,-6,4,-1},{1,5,25,77},50] (* Harvey P. Dale, Nov 22 2011 *)

a(n)=4*n*(1+2*n^2)/3+1 \\ Charles R Greathouse IV, Jul 07 2011

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
First differences: a(n+1) - a(n) = A108099(n).
Second differences: a(n+2) - 2*a(n+1) + a(n) = A008598(n+1).
Third differences: a(n+3) - 3*a(n+2) + 3*a(n+1) - a(n) = 16.
a(n) = (A168574(n) + A168547(n))/2. - This formula is the link to the Janet table of the PSE.
G.f.: ( 1 + x + 11*x^2 + 3*x^3 ) / (x-1)^4. - R. J. Mathar, Jul 07 2011
E.g.f.: (3 +12*x +24*x^2 +8*x^3)*exp(x)/3. - G. C. Greubel, Nov 02 2018

A181599 Numbers m with divisor 16 | m and abundance sigma(m)-2*m = 16.

Original entry on oeis.org

1504, 30592, 4526272, 8353792, 361702144, 1081850752, 1845991216, 2146926592, 21818579968, 34357510144, 228354264064, 549746900992, 2169800814592, 8796057370624, 24038405705152, 80952364306432, 140737345748992, 2737658648639872, 23810602502029312, 36979953305070592

Offset: 1

Vladimir Shevelev, Nov 01 2010

nonn

Cf. A097498, A141547, A181595, A181596, A181597, A181598, A118372, A045768, A000396, A005101, A153501, A005820.

A008598 INTERSECT A141547. - R. J. Mathar, Nov 04 2010

Definition rephrased - R. J. Mathar, Nov 04 2010
a(9)-a(13) from Donovan Johnson, Dec 08 2011
a(14)-a(20) from the b-file at A141547 added by Amiram Eldar, Aug 03 2024

A212950 Amounts (in cents) of Canadian coins in denominations suggested by Shallit.

Original entry on oeis.org

1, 5, 10, 25, 83, 100, 200

Offset: 1

Jonathan Vos Post, May 31 2012

			1c, 5c, 10c, 25c, 100c (a dollar coin, popularly known as a "loonie," because it bears a picture of a loon), 200c (the "toonie"), and the optimal suggested new coin in the denomination 83c.

Alan Burdick, The Physics of ... Pocket Change, Discover, October 2003 issue; published online October 1, 2003.

Cf. A212773, A212774, A008588, A008593, A008597, A008598, A135631.

Cf. A208953 (analog for American coins).

A327916 Triangle T(k, n) read by rows: Array A(k, n) = 2^k(k + 1 + 2n), k >= 0, n >= 0, read by antidiagonals upwards.

Original entry on oeis.org

1, 4, 3, 12, 8, 5, 32, 20, 12, 7, 80, 48, 28, 16, 9, 192, 112, 64, 36, 20, 11, 448, 256, 144, 80, 44, 24, 13, 1024, 576, 320, 176, 96, 52, 28, 15, 2304, 1280, 704, 384, 208, 112, 60, 32, 17, 5120, 2816, 1536, 832, 448, 240, 128, 68, 36, 19, 11264, 6144, 3328, 1792, 960, 512, 272, 144, 76, 40, 21

Offset: 0

Wolfdieter Lang, Oct 03 2019

The array A(k, n) arises from the following Pascal-type triangles PTodd(k), k >= 0 based on the positive odd integers A005408.
For example, the Pascal-type triangle PTodd(k), for k = 3 is
   1     3     5     7
      4     8    12
        12    20
           32
Taken upside-down such triangles become so-called addition towers of height k+1 (Rechenturm in German elementary schools; thanks to my correspondent Bennet D.), starting with any k+1 numbers. Here the positive odd numbers are used.
The sequence s of the final number of these Pascal-type triangles PT(k), for k >= 0, begins 1, 4, 12, 32, ...; s(k) = (k+1)*2^k = A001787(k+1), for k >= 0.
For k -> infinity the left-aligned row sequences build the array A(k, n), with k >= 0 and n >= 0, namely A(k, n) = 2^k*(k + 2*n + 1); this array begins:
  k\n  0   1   2   3   4    5 ...
  -------------------------------
  0:   1   3   5   7   9   11 ... {A005408(n)}
  1:   4   8  12  16  20   24 ... {A008586(n+1)}
  2:  12  20  28  36  44   52 ... {A017113(n+1)}
  3:  32  48  64  80  96  112 ... {A008598(n+2)}
  4:  80 112 144 176 208  240 ... {16*A005408(n+2)}
  5: 192 256 320 384 448  512 ... {A152691(n+3)}
  6: 448 576 704 832 960 1088 ... {64*A005408(n+3)}
  ...
The sequence s, the first (n=0) column of A, is always the binomial transform of the first (k=0) row in A.
A(k, n) = Sum_{j=0..k} binomial(k, j)*(2*(n+j)+1) = 2^k*(k + 1 + 2*n), for k >= 0 and n >= 0.
The corresponding antidiagonal-upwards read triangle is T(k, n) = A(k-n, n) = 2^(k-n)*(k + n + 1), n >= 0, k = 0..n.
If the nonnegative integers A001477 are used as k = 0 row of the array Anneg(k, n) = 2^(k-1)*(2*n + k), for k >= 0, n >= 0, with the triangle Tnneg(k, n) = Anneg(k-n, n) = (n + k)*2^(k-n-1), k >= 0, n = 0..k, then the s sequence is snneg(k) = Tnneg(k, 0) = k*2^{k-1} = A001787(k), the binomial transform of the sequence{A001477(n)}_{n>=0}. The triangle Tnneg begins [0], [1, 1], [4, 3, 2], [12, 8, 5, 3], [32, 20, 12, 7, 4], ... . See A062111 and the row-reversed triangle A152920 for other versions.

			The triangle T(k, n) begins:
   k\n    0    1    2    3   4   5   6   7  8  9 10 ...
  -----------------------------------------------------
   0:     1
   1:     4    3
   2:    12    8    5
   3:    32   20   12    7
   4:    80   48   28   16   9
   5:   192  112   64   36  20  11
   6:   448  256  144   80  44  24  13
   7:  1024  576  320  176  96  52  28  15
   8:  2304 1280  704  384 208 112  60  32 17
   9:  5120 2816 1536  832 448 240 128  68 36 19
  10: 11264 6144 3328 1792 960 512 272 144 76 40 21
  ...

Column sequences without leading zeros are for n=0..9: A001787(n+1), A001792(n+1), A045623(n+2), A045891(n+3), A034007(n+4), A111297(n+3), A159694(n+1), A159695(n+1), A159696(n+1), A159697(n+1).
The sequence of (sub)diagonal k, for k >= 0, is the row k sequence of array A: {(k + 2*n + 1)*2^k}_{k >= 0}.
Row sums: A213569(k+1), k >= 0 (see the J. M. Bergot comments there).
Cf. A006211, A152920.

Table[2^#*(# + 1 + 2 n) &[k - n], {k, 0, 10}, {n, 0, k}] // Flatten (* Michael De Vlieger, Oct 03 2019 *)

Array A(k, n) = Sum_{j=0..k} binomial(k, j)*(2*(n+j) + 1) = 2^k*(k + 1+ 2*n), for k >= 0 and n >= 0.
Triangle T(k, n) =  A(k-n, n) = 2^(k-n)*(k + n + 1), n >= 0, k = 0..n.
Recurrence: T(k, 0) = (k+1)*2^k = A001787(k+1), for k >= 0, and  T(k, n) = T(k, n-1) - T(k-1, n-1), for n >= 1, k >= 1, with T(k, n) = 0 if k < n.
O.g.f. for row polynomials: G(z,x) = Sum_{n=0..k} R(k, x)*z^n =
(1  + x*z*(1 - 4*z))/((1 - 2*z)^2*(1 - x*z)^2).
T(k, 0) = Sum_{n=0..k} binomial(k,n)*T(n, n), k >= 0 (binomial transform).

Definition corrected by Georg Fischer, Jul 13 2023

A212951 Amounts (in hundredths of a Euro) of coins in denominations suggested by Shallit.

Original entry on oeis.org

1, 2, 5, 10, 20, 50, 100, 133, 200

Offset: 1

Jonathan Vos Post, May 31 2012

The European Union uses eight coins - worth 1, 2, 5, 10, 20, and 50 cents, plus 1- and 2-Euro coins - with a range of values from 0 to 499. The average cost of making change in Europe, Jeffrey Shallit calculates, is 4.6 coins. The best way to lower the cost, to 3.92, would be for Europeans to add yet another coin, worth either 1.33 or 1.37 Euros (the sequence as shown uses 133, though 137 is an equally valid solution).

			1, 2, 5, 10, 20, and 50 cents, plus 1- and 2-Euro coins (100 and 200 cents), and the proposed 1.33-Euro coin (133 cents).

Alan Burdick, The Physics of ... Pocket Change. What we really need to solve the problem of loose and useless lucre is a new coin, Discover, October 2003 issue; published online October 1, 2003.

Cf. A212773, A212774, A008588, A008593, A008597, A008598, A135631.
Cf. A208953 (analog for American coins).
Cf. A212950 (analog for Canadian coins).

A363436 Array read by ascending antidiagonals: A(n, k) = k*n^2, with k >= 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 4, 2, 0, 0, 9, 8, 3, 0, 0, 16, 18, 12, 4, 0, 0, 25, 32, 27, 16, 5, 0, 0, 36, 50, 48, 36, 20, 6, 0, 0, 49, 72, 75, 64, 45, 24, 7, 0, 0, 64, 98, 108, 100, 80, 54, 28, 8, 0, 0, 81, 128, 147, 144, 125, 96, 63, 32, 9, 0, 0, 100, 162, 192, 196, 180, 150, 112, 72, 36, 10, 0

Offset: 0

Stefano Spezia, Jul 08 2023

			The array begins:
  0,  0,  0,   0,   0,   0,   0, ...
  0,  1,  2,   3,   4,   5,   6, ...
  0,  4,  8,  12,  16,  20,  24, ...
  0,  9, 18,  27,  36,  45,  54, ...
  0, 16, 32,  48,  64,  80,  96, ...
  0, 25, 50,  75, 100, 125, 150, ...
  0, 36, 72, 108, 144, 180, 216, ...
  ...

Cf. A000290 (k = 1), A001105 (k = 2), A033428 (k = 3), A016742 (k = 4), A033429 (k = 5), A033581 (k = 6), A033582 (k = 7), A139098 (k = 8), A016766 (k = 9), A033583 (k = 10), A033584 (k = 11), A135453 (k = 12),  A152742 (k = 13), A144555 (k = 14), A064761 (k = 15), A016802 (k = 16), A244630 (k = 17), A195321 (k = 18), A244631 (k = 19), A195322 (k = 20), A064762 (k = 21), A195323 (k = 22), A244632 (k = 23), A195824 (k = 24), A016850 (k = 25), A244633 (k = 26), A244634 (k = 27), A064763 (k = 28), A244635 (k = 29), A244636 (k = 30).
Cf. A001477 (n = 1), A008586 (n = 2), A008591 (n = 3), A008598 (n = 4), A008607 (n = 5), A044102 (n = 6), A152691 (n = 8).
Cf. A000007 (n = 0 or k = 0), A000578 (main diagonal), A002415 (antidiagonal sums), A004247.

A[n_,k_]:=k n^2; Table[A[n-k,k],{n,0,11},{k,0,n}]//Flatten

O.g.f.: x*y*(1 + x)/((1 - x)^3*(1 - y)^2).
E.g.f.: x*y*(1 + x)*exp(x + y).
A(n, k) = n*A004247(n, k).

cp's OEIS Frontend

A334566 Number of solutions of the Diophantine equation z^2 - y^2 - x^2 = n > 0 when the positive integers, x, y and z, are consecutive terms of an arithmetic progression.

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A351381 Table read by downward antidiagonals: T(n,k) = n*(k+1)^2.

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A171272 a(n) = 1 + 4*n*(1 + 2*n^2)/3.

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A181599 Numbers m with divisor 16 | m and abundance sigma(m)-2*m = 16.

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A212950 Amounts (in cents) of Canadian coins in denominations suggested by Shallit.

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A327916 Triangle T(k, n) read by rows: Array A(k, n) = 2^k*(k + 1 + 2*n), k >= 0, n >= 0, read by antidiagonals upwards.

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A212951 Amounts (in hundredths of a Euro) of coins in denominations suggested by Shallit.

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A363436 Array read by ascending antidiagonals: A(n, k) = k*n^2, with k >= 0.

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A171272 a(n) = 1 + 4n(1 + 2*n^2)/3.

A327916 Triangle T(k, n) read by rows: Array A(k, n) = 2^k(k + 1 + 2n), k >= 0, n >= 0, read by antidiagonals upwards.