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 11-14 of 14 results.

A161952 Base-15 Armstrong or narcissistic numbers (written in base 10).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 113, 128, 2755, 3052, 5059, 49074, 49089, 386862, 413951, 517902, 15219156, 18605333, 38009273, 40082196, 40310423, 40868227, 47527794, 100128060, 100128061, 100128188, 104189152, 105464820
Offset: 1

Views

Author

Joseph Myers, Jun 22 2009

Keywords

Comments

Whenever 15|a(n) (n = 32, 36, 40, 86, 100, 135, 143, 194, 197, 201), then a(n+1) = a(n) + 1. Zero also satisfies the definition (n = Sum_{i=1..k} d[i]^k where d[1..k] are the base-15 digits of n), but this sequence only considers positive terms. - M. F. Hasler, Nov 22 2019

Crossrefs

In other bases: A010344 (base 4), A010346 (base 5), A010348 (base 6), A010350 (base 7), A010354 (base 8), A010353 (base 9), A005188 (base 10), A161948 (base 11), A161949 (base 12), A161950 (base 13), A161951 (base 14), A161953 (base 16).

Programs

  • Mathematica
    Select[Range[10^7], # == Total[IntegerDigits[#, 15]^IntegerLength[#, 15]] &] (* Michael De Vlieger, Nov 04 2020 *)
  • PARI
    select( is_A161952(n)={n==vecsum([d^#n|d<-n=digits(n,15)])}, [1..10^5]) \\ M. F. Hasler, Nov 22 2019

Extensions

Terms sorted in increasing order by Pontus von Brömssen, Mar 03 2019

A162234 Base 9 perfect digital invariants (written in base 10): numbers equal to the sum of the k-th powers of their base-9 digits, for some k.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 27, 28, 41, 50, 126, 127, 243, 244, 353, 468, 469, 1052, 1824, 2187, 2188, 8052, 8295, 9857, 19683, 19684, 36804, 65538, 65539, 177147, 177148, 1198372, 1594323, 1594324, 3357009, 3357010, 5300099, 6287267, 10097892
Offset: 1

Views

Author

Joseph Myers, Jun 28 2009

Keywords

Comments

Whenever a(n) is a multiple of 9, then a(n+1) = a(n) + 1 is also a base 9 perfect digital invariant, with the same exponent k. - M. F. Hasler, Nov 21 2019

Crossrefs

Cf. A162235 (corresponding exponents), A010353 (restriction to power = number of digits), A033841, A162236. In other bases: A162216 (base 3), A162219 (base 4), A162222 (base 5), A162225 (base 6), A162228 (base 7), A162231 (base 8), A023052 (base 10).

Programs

  • PARI
    select( {is_A162234(n, b=9)=nn|| return(t==n))}, [0..10^5]) \\ M. F. Hasler, Nov 21 2019

A010352 Base-9 Armstrong or narcissistic numbers, written in base 9.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 45, 55, 150, 151, 570, 571, 2446, 12036, 12336, 14462, 2225764, 6275850, 6275851, 12742452, 356614800, 356614801, 1033366170, 1033366171, 1455770342, 8463825582, 131057577510, 131057577511
Offset: 1

Views

Author

Keywords

Comments

From M. F. Hasler, Nov 18 2019: (Start)
Like the other single-digit terms, zero would satisfy the definition (n = Sum_{i=1..k} d[i]^k when d[1..k] are the base-9 digits of n), but here only positive numbers are considered.
Terms a(n+1) = a(n) + 1 (n = 11, 13, 20, 23, 25, 29, 33, 48, 51, 57) correspond to solutions a(n) that are multiples of 9, in which case a(n) + 1 is also a solution. (End)

Examples

			126 = 150_9 (= 1*9^2 + 5*9^1 + 0*9^0) = 1^3 + 5^3 + 0^3. It is easy to see that 126 + 1 then also satisfies this relation, as for all other terms that are multiples of 9. - _M. F. Hasler_, Nov 21 2019
		

Crossrefs

Cf. A010353 (a(n) written in base 10).
In other bases: A010343 (base 4), A010345 (base 5), A010347 (base 6), A010349 (base 7), A010351 (base 8), A005188 (base 10).

Programs

Extensions

Edited by Joseph Myers, Jun 28 2009

A351374 Base-20 Armstrong or narcissistic numbers (written in base 10).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2413, 53808, 760400, 760401, 45661018, 62470211, 619939142, 14613048357, 1421043363262183, 48470736648305918, 514822672411130775, 360672575087017687943, 264237343348909655564587, 267218514330351511200145
Offset: 1

Views

Author

Giovanni Corbelli, Mar 18 2022

Keywords

Comments

Written in base twenty the numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J, 60D, 6EA8, 4F100, 4F101, E57CAI, JA8FAB, 9DEC7H2, B86BB0HH.
Sequence is finite. Since k*19^k < 20^(k-1) for k >= 157, all terms must have less than 157 base-20 digits. 20*m is a term if and only if 20*m+1 is a term. - Chai Wah Wu, Apr 20 2022

Examples

			2413 is in the sequence because 2413 is 60D in base 20 (D stands for 13) and 6^3 + 0^3 + 13^3 = 2413. (The exponent 3 is the number of base-20 digits.)
		

Crossrefs

In other bases: A010344 (base 4), A010346 (base 5), A010348 (base 6), A010350 (base 7), A010354 (base 8), A010353 (base 9), A005188 (base 10), A161948 (base 11), A161949 (base 12), A161950 (base 13), A161951 (base 14), A161952 (base 15), A161953 (base 16).

Programs

  • Mathematica
    Select[Range[10^6], # == Total[ IntegerDigits[#, 20]^IntegerLength[#, 20]] &]
  • PARI
    isok(m) = my(d=digits(m, 20)); sum(k=1, #d, d[k]^#d) == m; \\ Michel Marcus, Mar 19 2022
    
  • Python
    from itertools import islice, combinations_with_replacement
    from sympy.ntheory.factor_ import digits
    def A351374_gen(): # generator of terms
        for k in range(1,157):
            a = tuple(i**k for i in range(20))
            yield from (x[0] for x in sorted(filter(lambda x:x[0] > 0 and tuple(sorted(digits(x[0],20)[1:])) == x[1], \
                              ((sum(map(lambda y:a[y],b)),b) for b in combinations_with_replacement(range(20),k)))))
    A351374_list = list(islice(A351374_gen(),20)) # Chai Wah Wu, Apr 20 2022

Extensions

a(28)-a(30) from Chai Wah Wu, Apr 20 2022
a(31)-a(33) from Jinyuan Wang, May 05 2025
Previous Showing 11-14 of 14 results.