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.

Previous Showing 11-18 of 18 results.

A174139 Numbers congruent to {0,1,2,3,4,10,11,12,13,14,20,21,22,23,24} mod 25.

Original entry on oeis.org

0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 85, 86, 87, 88, 89, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 110, 111, 112
Offset: 1

Views

Author

Rick L. Shepherd, Mar 09 2010

Keywords

Comments

Numbers whose partition into parts of sizes 1, 5, 10, and 25 having a minimal number of parts does not include a part of size 5.
For each number the partition is unique.
Complement of A174138.
Amounts in cents not including a nickel when the minimal number of coins is selected from pennies, nickels, dimes, and quarters (whether usage of bills for whole-dollar amounts is permitted or not).
For each n >= 0, floor(n/25) parts of size 25 (quarters) occur in the partition with minimal number of these parts (regardless of whether partition includes part of size 5).
First differs from A032955 at n = 76. - Avi Mehra, Oct 08 2020

Crossrefs

Cf. A174138, A174140, A174141, A047201 (requires at least one part of size 1 (penny)), A008587, A053344 (minimal number of parts), A001299 (number of all such partitions).

Programs

  • Mathematica
    Select[Range[0, 112], Mod[Mod[#, 25], 10] < 5 &] (* Amiram Eldar, Oct 08 2020 *)
  • PARI
    { my(table=[0,1,2,3,4, 10,11,12,13,14, 20,21,22,23,24]);
    a(n) = my(r);[n,r]=divrem(n-1,15); 25*n + table[r+1]; } \\ Kevin Ryde, Oct 08 2020

Formula

a(15+n) = a(n) + 25 for n >= 1.
From R. J. Mathar, Oct 08 2011: (Start)
a(n) = +a(n-1) +a(n-15) -a(n-16).
G.f.: x^2*(1 +x +x^2 +x^3 +6*x^4 +x^5 +x^6 +x^7 +x^8 +6*x^9 +x^10 +x^11 +x^12 +x^13+x^14) / ( (1+x+x^2) *(x^4+x^3+x^2+x+1) *(x^8-x^7+x^5-x^4+x^3-x+1) *(x-1)^2). (End)

A208216 Number of ways of making change for n cents using coins of 1, 5, 10, and 25 cents, where each coin that appears is used a different number of times.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 4, 3, 3, 4, 4, 5, 5, 5, 4, 5, 8, 7, 6, 8, 7, 12, 9, 11, 11, 12, 14, 14, 15, 14, 14, 20, 16, 18, 18, 20, 25, 24, 23, 24, 24, 32, 27, 30, 31, 30, 39, 38, 36, 39, 37, 47, 43, 47, 45, 48, 55, 53, 53, 56, 57, 69, 61, 65, 65, 67, 78, 77
Offset: 0

Views

Author

Matthew C. Russell, Apr 24 2012

Keywords

Examples

			For n=6 the a(6)=1 solution is six copies of the 1-cent coin. (Taking one 1-cent coin and one 5-cent coin is not allowed, as those two coins have the same nonzero multiplicity.)
		

Crossrefs

Programs

  • Maple
    # for a g.f. in Maple format see the Doron Zeilberger link.

A057536 Minimal number of coins needed to pay n Euro-cents using the Euro currency.

Original entry on oeis.org

0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 3, 4, 4, 5, 5, 4, 5, 5, 6, 6, 3, 4, 4, 5, 5, 4, 5, 5, 6, 6, 1, 2, 2, 3, 3, 2
Offset: 0

Views

Author

Thomas Brendan Murphy (murphybt(AT)tcd.ie), Sep 06 2000

Keywords

Comments

Euro currency has coins and bills of size 1, 2, 5, 10, 20, 50, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000 cents.

Examples

			a(57) = 3 because to pay 57 cents at least 3 coins are needed: 1 of 50 cents, 1 of 5 cents and 1 of 2 cents.
		

Crossrefs

Programs

  • Mathematica
    Table[Min[Map[Total,FrobeniusSolve[{1,2,5,10,20,50,100,200,500,1000,2000,5000,10000,20000,50000},n]]],{n, 0, 105}] (* Joan Ludevid, Jun 15 2022 *)
    numCoins[n_]:=(amount = n; coins = {50000, 20000, 10000, 5000, 2000, 1000, 500, 200, 100, 50, 20, 10, 5, 2, 1}; total=0; For[i = 1, i <= Length[coins], i++, total+=Quotient[amount, coins[[i]]]; amount = Mod[amount, coins[[i]]]]; total);
    Table[numCoins[n], {n, 0, 105}] (* Joan Ludevid, Jun 16 2022 *)

Formula

a(n) = floor(n/50000) + floor((n mod 50000)/20000) + floor(((n mod 50000) mod 20000)/10000) + ... + floor(((n mod 50000) mod 20000 ... mod 5)/2) + ((n mod 50000) mod 20000)... mod 2.

Extensions

a(0)=0 prepended by Alois P. Heinz, May 26 2015

A351740 Total number of parts in all partitions of n into parts of size 1, 5, 10 or 25.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 8, 10, 12, 14, 19, 23, 27, 31, 35, 44, 50, 56, 62, 68, 83, 92, 101, 110, 119, 141, 154, 167, 180, 193, 226, 244, 262, 280, 298, 343, 367, 391, 415, 439, 500, 531, 562, 593, 624, 702, 741, 780, 819, 858, 959, 1008, 1057, 1106, 1155, 1280, 1340
Offset: 0

Views

Author

Alois P. Heinz, Feb 17 2022

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0 or i=1, [1, n], b(n, i-1)+
         (p->`if`(p>n, 0, (t->t+[0, t[1]])(b(n-p, i))))([1, 5, 10, 25][i]))
        end:
    a:= n-> b(n, 4)[2]:
    seq(a(n), n=0..100);

Formula

a(n) = Sum_{k=0..n} k * A351725(n,k).

A351724 Number of compositions of n into parts of size 1, 5, 10 or 25.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 9, 13, 18, 24, 31, 42, 58, 80, 109, 146, 197, 268, 366, 499, 676, 916, 1243, 1690, 2299, 3122, 4237, 5751, 7811, 10614, 14418, 19580, 26587, 36106, 49043, 66614, 90473, 122869, 166866, 226632, 307810, 418060, 567784, 771122, 1047296, 1422396, 1931845
Offset: 0

Views

Author

R. J. Mathar, Feb 17 2022

Keywords

Comments

Starts to differ from A114044 at n=50.

Examples

			a(8)=5 counts 5 compositions 1+1+1+1+1+1+1+1 = 1+1+1+5 = 1+1+5+1 = 1+5+1+1 = 5+1+1+1.
		

Crossrefs

Cf. A114044 (parts 50 and 100 admitted), A001299 (partitions).
Row sums of A351726.

Formula

G.f. : 1/(1-x-x^5-x^10-x^25).
a(n) = +a(n-1) +a(n-5) +a(n-10) +a(n-25).

A129263 Skylar (age 7) counts change by stacking all coins of the same type then arranging the stacks in a row. a(n) is the number of distinct Skylar stackings of n cents using any combination of pennies, nickels, dimes or quarters.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 3, 3, 3, 3, 5, 7, 7, 7, 7, 10, 15, 15, 15, 15, 19, 25, 25, 25, 25, 31, 41, 41, 41, 41, 49, 63, 63, 63, 63, 74, 95, 95, 95, 95, 111, 147, 147, 147, 147, 166, 209, 209, 209, 209, 234, 293, 293, 293, 293, 322, 391, 391, 391, 391, 427, 515, 515, 515, 515
Offset: 0

Views

Author

Andrew V. Sutherland, Aug 20 2007

Keywords

Comments

Sequence definition and Scratch program to compute the 100 terms due to Skylar Sutherland. Generating function contributed by Andrew V. Sutherland. Related to A001299, but distinguishes permutations of coin types.

Examples

			a(16) = 15 = 1+2*4+6*1 since the distinct Skylar stackings of 16 cents are:
16p, 11p1n, 1n11p, 6p2n, 2n6p, 1p3n, 3n1p, 1p1d, 1d1p, 1p1n1d, 1p1d1n, 1n1p1d, 1n1d1p, 1d1p1n, 1d1n1p
		

References

  • Skylar Sutherland, student presentation at "The Undiscovered Country", a course for young mathematicians. Part of MIT's Educational Studies Program.

Crossrefs

Cf. A001299.

Formula

Let A_v(x,y) = 1-y+y/(1-x)^v and A(x,y) = A_1(x,y)A_5(x,y)A_10(x,y)A_25(x,y). Let A^(k)(x,y) denote the k-th partial derivative of A(x,y) w.r.t. y. The generating function of a(n) is A(x) = Sum A^(k)(x,0) for k from 0 to 4.

A245574 Minimal coin changing sequence for denominations 1, 2, 5, 10, 20 and 50 cents.

Original entry on oeis.org

0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 3, 4, 4, 5, 5, 4, 5, 5, 6, 6, 3, 4, 4, 5, 5, 4, 5, 5, 6, 6, 2
Offset: 0

Views

Author

Marko Riedel, Jul 25 2014

Keywords

Comments

a(100) = 2 is the first term where the sequence is different from A057536. - Colin Barker, Jul 26 2014

Crossrefs

Extensions

More terms from Colin Barker, Jul 26 2014
a(0)=0 prepended by Alois P. Heinz, May 26 2015

A339094 Number of (unordered) ways of making change for n US Dollars using the current US denominations of $1, $2, $5, $10, $20, $50 and $100 bills.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 5, 6, 7, 8, 11, 12, 15, 16, 19, 22, 25, 28, 31, 34, 41, 44, 51, 54, 61, 68, 75, 82, 89, 96, 109, 116, 129, 136, 149, 162, 175, 188, 201, 214, 236, 249, 271, 284, 306, 328, 350, 372, 394, 416, 451, 473, 508, 530, 565, 600, 635, 670, 705, 740, 793, 828, 881, 916
Offset: 0

Views

Author

Robert G. Wilson v, Nov 25 2020

Keywords

Comments

Not the same as A001313. First difference appears at A001313(100) being 4562, whereas a(100) is 4563; obviously one more than A001313(100).
Not the same as A057537.
Number of partitions of n into parts 1, 2, 5, 10, 20, 50 and 100.

Examples

			a(5) is 4 because 1+1+1+1+1 = 2+1+1+1 = 2+2+1 = 5.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Length@ IntegerPartitions[n, All, {1, 2, 5, 10, 20, 50, 100}]; Array[f, 75, 0] (* or *)
    CoefficientList[ Series[1/((1 - x) (1 - x^2) (1 - x^5) (1 - x^10) (1 - x^20) (1 - x^50) (1 - x^100)), {x, 0, 75}], x] (* or *)
    Table[ Length@ FrobeniusSolve[{1, 2, 5, 10, 20, 50, 100}, n], {n, 0, 75}] (* much slower *)
  • PARI
    coins(v[..])=my(x='x); prod(i=1, #v, 1/(1-x^v[i]))
    Vec(coins(1, 2, 5, 10, 20, 50, 100)+O(x^99)) \\ Charles R Greathouse IV, Jan 24 2022

Formula

G.f.: 1/((1-x)*(1-x^2)*(1-x^5)*(1-x^10)*(1-x^20)*(1-x^50)*(1-x^100)).
Previous Showing 11-18 of 18 results.