A005188 -id:A005188 - cp's OEIS frontend

A101337 Sum of (each digit of n raised to the power (number of digits in n)).

1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 5, 10, 17, 26, 37, 50, 65, 82, 4, 5, 8, 13, 20, 29, 40, 53, 68, 85, 9, 10, 13, 18, 25, 34, 45, 58, 73, 90, 16, 17, 20, 25, 32, 41, 52, 65, 80, 97, 25, 26, 29, 34, 41, 50, 61, 74, 89, 106, 36, 37, 40, 45, 52, 61, 72, 85, 100, 117, 49, 50, 53, 58, 65

Offset: 1

Gordon Hamilton, Dec 24 2004

Sometimes referred to as "narcissistic function" (in base 10). Fixed points are the narcissistic (or Armstrong, or plus perfect) numbers A005188. - M. F. Hasler, Nov 17 2019

			a(75) = 7^2 + 5^2 = 74 and a(705) = 7^3 + 0^3 + 5^3 = 468.
a(1.02e59 - 1) = 102429587095122578993551250282047487264694110769657513064859 ~ 1.024e59 is an example of n close to the limit beyond which a(n) < n for all n. - _M. F. Hasler_, Nov 17 2019

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Wikipedia, Narcissistic number, as of Nov 18 2019.

Cf. A055642, A179239, A306360.

f:=func; [f(n):n in [1..75]]; // Marius A. Burtea, Nov 18 2019

Array[Total[IntegerDigits[#]^IntegerLength[#]]&,80] (* Harvey P. Dale, Aug 27 2011 *)

a(n)=my(d=digits(n)); sum(i=1,#d, d[i]^#d) \\ Charles R Greathouse IV, Aug 10 2017

apply( A101337(n)=vecsum([d^#n|d<-n=digits(n)]), [0..99]) \\ M. F. Hasler, Nov 17 2019

def A101337(n):
    s = str(n)
    l = len(s)
    return sum(int(d)**l for d in s) # Chai Wah Wu, Feb 26 2019

a(n) <= A055642(n)*9^A055642(n) with equality for all n = 10^k - 1. Write n = 10^x to get a(n) < n when 1+log_10(x+1) < (x+1)(1-log_10(9)) <=> x > 59.85. It appears that a(n) < n already for all n > 1.02*10^59. - M. F. Hasler, Nov 17 2019

Name changed by Axel Harvey, Dec 26 2011

Edited by M. F. Hasler, Nov 17 2019

A093137 Expansion of (1-7x)/((1-x)(1-10*x)).

Original entry on oeis.org

1, 4, 34, 334, 3334, 33334, 333334, 3333334, 33333334, 333333334, 3333333334, 33333333334, 333333333334, 3333333333334, 33333333333334, 333333333333334, 3333333333333334, 33333333333333334, 333333333333333334, 3333333333333333334, 33333333333333333334

Offset: 0

Paul Barry, Mar 24 2004

Second binomial transform of 3*A001045(3n)/3+(-1)^n. Partial sums of A093138. A convex combination of 10^n and 1. In general the second binomial transform of k*Jacobsthal(3n)/3+(-1)^n is 1,1+k,1+11k,1+111k,... This is the case for k=3.
a(n) is the number of n-length sequences of decimal digits whose sum is divisible by 3. - Geoffrey Critzer, Jan 18 2014
This sequence appears in a family of curious cubic identities based on the Armstrong number 407 = A005188(13). See the formula section. For the analog identities based on 153 = A005188(10) see a comment on A246057 with the van der Poorten et al. reference and A281857. For those based on 370 = A005188(11) see A067275, A002277 and A281858. - Wolfdieter Lang, Feb 08 2017

			a(1)^2 = 16
a(2)^2 = 1156
a(3)^2 = 111556
a(4)^2 = 11115556
a(5)^2 = 1111155556
a(6)^2 = 111111555556
a(7)^2 = 11111115555556
a(8)^2 = 1111111155555556
a(9)^2 = 111111111555555556, etc... (see A102807). - _Philippe Deléham_, Oct 03 2011
Curious cubic identities: 407 = 4^3 + 0^3 + 7^3, 340067 = 34^3 + (00)^3 + 67^3, 334000677 = 334^3 + (000)^3 + 677^3, ... - _Wolfdieter Lang_, Feb 08 2017

David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See entry 3334 at p. 168.

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A005188, A067275, A246057, A281857, A281859.
Cf. A001045, A002277, A093138, A281858.
Cf. A102807 (squared).

nn=20; r=Solve[{s==4x s+3 x a+3x b+1,a==4x a+3x s+3x b,b==4x b+3x s+3x a},{s,a,b}]; CoefficientList[Series[s/.r,{x,0,nn}],x] (* Geoffrey Critzer, Jan 18 2014 *)
Table[3*10^n/9 + 6/9, {n, 0, 20}] (* or *) NestList[10 # - 6 &, 1, 20] (* Michael De Vlieger, Feb 08 2017 *)
LinearRecurrence[{11,-10},{1,4},20] (* Harvey P. Dale, Oct 07 2017 *)

Vec((1-7*x)/((1-x)*(1-10*x)) + O (x^30)) \\ Michel Marcus, Feb 09 2017

a(n) = 3*10^n/9 + 6/9.
a(n) = 10*a(n-1)-6 with a(0)=1. - Vincenzo Librandi, Aug 02 2010
a(n)^3 + 0(n)^3 + A067275(n+1)^3 = concatenation(a(n), 0(n), A067275(n+1)) = A281859(n), where 0(n) denotes n 0's, n >= 1. - Wolfdieter Lang, Feb 08 2017
From Elmo R. Oliveira, Aug 17 2024: (Start)
E.g.f.: exp(x)*(exp(9*x) + 2)/3.
a(n) = 11*a(n-1) - 10*a(n-2) for n > 1. (End)

A010346 Base-5 Armstrong or narcissistic numbers (written in base 10).

Original entry on oeis.org

1, 2, 3, 4, 13, 18, 28, 118, 289, 353, 419, 4890, 4891, 9113, 1874374, 338749352, 2415951874

Offset: 1

N. J. A. Sloane

Zero would also satisfy the definition as the other single-digit terms, but here only positive numbers are considered. - M. F. Hasler, Nov 20 2019

Eric Weisstein's World of Mathematics, Narcissistic Number
D. T. Winter, Table of Armstrong Numbers

Cf. A010345 (a(n) written in base 5).

In other bases: A010344 (base 4), 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).

A010346=select( is_A010346(n)={n==vecsum([d^#n|d<-n=digits(n,5)])}, [0..9999]) \\ This yields only terms < 10^4 (i.e., all but the last 3 terms), for illustration of is_A010346(). In older versions of PARI, use {n==sum(i=1,#n=digits(n,5),n[i]^#n)}. - M. F. Hasler, Nov 20 2019

Edited by Joseph Myers, Jun 28 2009

A010354 Base-8 Armstrong or narcissistic numbers (written in base 10).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 20, 52, 92, 133, 307, 432, 433, 16819, 17864, 17865, 24583, 25639, 212419, 906298, 906426, 938811, 1122179, 2087646, 3821955, 13606405, 40695508, 423056951, 637339524, 6710775966, 13892162580, 32298119799, 97095152738, 98250308556, 98317417420, 125586038802

Offset: 1

N. J. A. Sloane

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 8 digits of n), but here only positive numbers are considered. - M. F. Hasler, Nov 20 2019

			From _M. F. Hasler_, Nov 20 2019: (Start)
20 = 24_8 (in base 8), and 2^2 + 4^2 = 20.
432 = 660_8, and 6^3 + 6^3 + 0^3 = 432; it's easy to see that 432 + 1 then also satisfies the equation, as for any term that is a multiple of 8. (End)

Joseph Myers, Table of n, a(n) for n = 1..62 (the full list of terms, from Winter)
Eric Weisstein's World of Mathematics, Narcissistic Number
D. T. Winter, Table of Armstrong Numbers

Cf. A010351 (a(n) written in base 8).

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

select( {is_A010354(n)=n==vecsum([d^#n|d<-n=digits(n,8)])}, [0..10^6]) \\ This gives only terms < 10^6, for illustration of is_A010354(). - M. F. Hasler, Nov 20 2019

from itertools import islice, combinations_with_replacement
def A010354_gen(): # generator of terms
    for k in range(1,30):
        a = tuple(i**k for i in range(8))
        yield from (x[0] for x in sorted(filter(lambda x:x[0] > 0 and tuple(int(d,8) for d in sorted(oct(x[0])[2:])) == x[1], \
                          ((sum(map(lambda y:a[y],b)),b) for b in combinations_with_replacement(range(8),k)))))
A010354_list = list(islice(A010354_gen(),20)) # Chai Wah Wu, Apr 20 2022

Edited by Joseph Myers, Jun 28 2009

A010344 Base-4 Armstrong or narcissistic numbers (written in base 10).

Original entry on oeis.org

1, 2, 3, 28, 29, 35, 43, 55, 62, 83, 243

Offset: 1

N. J. A. Sloane

Eric Weisstein's World of Mathematics, Narcissistic Number
D. T. Winter, Table of Armstrong Numbers

Cf. A010343 (a(n) written in base 4).

In other bases: 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).

A010344=select( n->n==vecsum([d^#n|d<-n=digits(n,4)]), [0..333]) \\ M. F. Hasler, Nov 18 2019

Edited by Joseph Myers, Jun 28 2009

A007532 Handsome numbers: sum of positive powers of its digits; a(n) = Sum_{i=1..k} d[i]^e[i] where d[1..k] are the decimal digits of a(n), e[i] > 0.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 24, 43, 63, 89, 132, 135, 153, 175, 209, 224, 226, 262, 264, 267, 283, 332, 333, 334, 357, 370, 371, 372, 373, 374, 375, 376, 377, 378, 379, 407, 445, 463, 518, 598, 629, 739, 794, 849, 935, 994, 1034

Offset: 1

N. J. A. Sloane, Robert G. Wilson v

The previous name was "Powerful numbers, Definition (2). Cf. A001694, A023052. - N. J. A. Sloane, Jan 16 2022

J. Randle has suggested the name "powerful numbers" for the perfect digital invariants A023052, equal to the sum of a fixed power of the digits. However, "powerful" usually refers to a prime factorization related property, cf. A001694 (and references there as well as on the MathWorld page). C. Rivera has suggested the name "handsome" for these numbers (in view of narcissistic numbers A005188) in his prime puzzle #15: see also contributed comments concerning terminology on that page. - M. F. Hasler, Nov 21 2019

			43 = 4^2 + 3^3 is OK; 254 = 2^7 + 5^3 + 4^0 is not OK since one of the powers is 0.

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

David W. Wilson, Table of n, a(n) for n = 1..10000
Giovanni Resta, d-powerful numbers, the 30067 terms and sums up to 10^6.
Carlos Rivera, Puzzle 15.- Narcissistic and Handsome Primes, The Prime Puzzles and Problems Connection.
Eric Weisstein's World of Mathematics, Powerful Number.
Index entries for sequences related to powerful numbers

Cf. A001694, A005934, A005188, A003321, A014576, A023052, A046074, A050240 (>= 2 reps.), A050241.

Different from A061862.

a007532 n = a007532_list !! (n-1)
a007532_list = filter f [1..] where
   f x = g x 0 where
     g 0 v = v == x
     g u v = if d <= 1 then g u' (v + d) else v <= x && h d
             where h p = p <= x && (g u' (v + p) || h (p * d))
                   (u', d) = divMod u 10
-- Reinhard Zumkeller, Jun 02 2013

N:= 10000; # to get all entries <= N
Sums:= proc(L,N)
  option remember;
  local x1,L1;
  x1:= L[1];
  if x1 = 1 then L1:= {1}
  else L1:= {seq(x1^j,j=1..floor(log[x1](N)))};
  fi;
  if nops(L) = 1 then L1
  else select(`<=`,{seq(seq(a+b,a=L1),b=Sums(L[2..-1],N))},N)
  fi
end proc;
filter:= proc(x,N)
   local L;
   L:= sort(subs(0=NULL,convert(x,base,10))) ;
   member(x, Sums(L,N));
end proc;
A007532:= select(filter,[$1..N],N); # Robert Israel, Apr 13 2014

Select[Range@1000,(s=#;MemberQ[Total/@(a^#&/@Tuples[Range@If[#==1||#==0,1,Floor[Log[#,s]]]&/@(a=IntegerDigits[s])]),s])&] (* Giorgos Kalogeropoulos, Aug 18 2021 *)

from itertools import count, takewhile
def cands(n, d):
    return takewhile(lambda x: x<=n, (d**i for i in count(1)))
def handsome(s, t):
    if s == "":
        return t == 0
    if s[0] in "01":
        return handsome(s[1:], t - int(s[0]))
    return any(handsome(s[1:], t - p) for p in cands(t, int(s[0])))
def ok(n):
    return n and handsome(str(n), n)
print(list(filter(ok, range(1035)))) # Michael S. Branicky, Aug 18 2021

If n = d_1 d_2 ... d_k in decimal, then there are integers m_1, m_2, ..., m_k > 0 such that n = d_1^m_1 + ... + d_k^m_k.

A010348 Base-6 Armstrong or narcissistic numbers (written in base 10).

Original entry on oeis.org

1, 2, 3, 4, 5, 99, 190, 2292, 2293, 2324, 3432, 3433, 6197, 36140, 269458, 391907, 10067135, 2510142206, 2511720147, 3866632806, 3866632807, 3930544834, 4953134588, 5018649129, 6170640875, 124246559501, 4595333541803, 5341093125744, 5341093125745, 19418246235419

Offset: 1

N. J. A. Sloane

From M. F. Hasler, Nov 20 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 digits of n), but here only positive numbers are considered.
Terms a(n+1) = a(n) + 1 (n = 8, 11, 20, 28) correspond to solutions a(n) ending in a digit 0 in base 6, in which case a(n) + 1 also is a solution. (End)

Joseph Myers, Table of n, a(n) for n = 1..30 (the full list of terms, from Winter)
Eric Weisstein's World of Mathematics, Narcissistic Number
D. T. Winter, Table of Armstrong Numbers (latest backup on web.archive.org from Jan. 2010; page no longer available), published not later than Aug. 2003.

Cf. A010347 (a(n) written in base 6).

In other bases: A010344 (base 4), A010346 (base 5), 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).

select( {is_A010348(n)=n==vecsum([d^#n|d<-n=digits(n,6)])}, [0..4e5\1]) \\ Note: this yields only terms < 10^6, for illustration of is_A010348(). - M. F. Hasler, Nov 20 2019

Edited by Joseph Myers, Jun 28 2009

A010350 Base-7 Armstrong or narcissistic numbers (written in base 10).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 10, 25, 32, 45, 133, 134, 152, 250, 3190, 3222, 3612, 3613, 4183, 9286, 35411, 191334, 193393, 376889, 535069, 794376, 8094840, 10883814, 16219922, 20496270, 32469576, 34403018, 416002778, 416352977, 420197083, 725781499, 1500022495, 15705029375, 15705029376, 28700208851

Offset: 1

N. J. A. Sloane

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 digits of n), but here only positive numbers are considered. - M. F. Hasler, Nov 20 2019

Joseph Myers, Table of n, a(n) for n = 1..59 (the full list of terms, from Winter)
Eric Weisstein's World of Mathematics, Narcissistic Number
D. T. Winter, Table of Armstrong Numbers

Cf. A010349 (a(n) written in base 7).

In other bases: A010344 (base 4), A010346 (base 5), A010348 (base 6), 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).

select( {is_A010350(n)=n==vecsum([d^#n|d<-n=digits(n,7)])}, [0..10^6]) \\ This yields only terms < 10^6, for illustration of is_A010350(). - M. F. Hasler, Nov 20 2019

Edited by Joseph Myers, Jun 28 2009

A010353 Base-9 Armstrong or narcissistic numbers (written in base 10).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 41, 50, 126, 127, 468, 469, 1824, 8052, 8295, 9857, 1198372, 3357009, 3357010, 6287267, 156608073, 156608074, 403584750, 403584751, 586638974, 3302332571, 42256814922, 42256814923, 114842637961, 155896317510, 552468844242, 552468844243, 647871937482, 686031429775

Offset: 1

N. J. A. Sloane

From M. F. Hasler, Nov 20 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)

			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 20 2019

Joseph Myers, Table of n, a(n) for n = 1..58 (the full list of terms, from Winter)
René-Louis Clerc, Perfect r-narcissistic numbers in any base, hal-04376934, 2024.
Eric Weisstein's World of Mathematics, Narcissistic Number
D. T. Winter, Table of Armstrong Numbers

Cf. A010352 (a(n) written in base 9).

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

Select[Range[9^7], # == Total[IntegerDigits[#, 9]^IntegerLength[#, 9]] &] (* Michael De Vlieger, Jan 17 2024 *)

select( {is_A010353(n)=n==vecsum([d^#n|d<-n=digits(n,9)])}, [0..10^4]) \\ This gives only terms < 10^6, for illustration of is_A010353(). - M. F. Hasler, Nov 20 2019

Edited by Joseph Myers, Jun 28 2009

A067275 Number of Fibonacci numbers A000045(k), k <= 10^n, which end in 4.

Original entry on oeis.org

0, 1, 7, 67, 667, 6667, 66667, 666667, 6666667, 66666667, 666666667, 6666666667, 66666666667, 666666666667, 6666666666667, 66666666666667, 666666666666667, 6666666666666667, 66666666666666667, 666666666666666667, 6666666666666666667, 66666666666666666667, 666666666666666666667

Offset: 0

Joseph L. Pe, Feb 21 2002

Numbers n such that the digits of A000326(n), the n-th pentagonal number, begin with n. [original definition of this sequence]
The sequence 1,7,67.... has a(n) = 6*10^n/9+3/9. It is the second binomial transform of 6*A001045(3n)/3 + (-1)^n. In general the second binomial transform of k*Jacobsthal(3n)/3+(-1)^n is k*10^n/9 + (1-k/9) = 1, 1+k, 1+11k, 1+111k, ... - Paul Barry, Mar 24 2004
Except for the first two terms, these are the 3-automorphic numbers ending in 7. - Eric M. Schmidt, Aug 28 2012
From Wolfdieter Lang, Feb 08 2017: (Start)
This sequence appears in curious identities based on the Armstrong numbers 370 = A005188(11), 371 = A005188(12) and 407 = A005188(13).
For such identities based on 153 = A005188(10) see a comment in A246057 with the van der Poorten et al. reference.
For 370 these identities are A002277(n)^3 + a(n+1)^3  + 0(n)^3 = A002277(n)*10^(2*n) + a(n+1)*10^n + 0(n) = A281858(n), where 0(n) means n 0's.
For 371 these identities are A002277(n)^3 + a(n+1)^3 + (0(n-1)1)^3 = A002277(n)*10^(2*n) + a(n+1)*10^n + 0(n-1)1 = A281860(n), where 0(n-1)1 means n-1 0's followed by a 1.
For 407 these identities are A093137(n)^3 + 0(n)^3 + a(n+1)^3 = concatenation(A093137(n), 0(n), a(n+1)) = A281859(n). (End)

			a(2) = 7 because 7 of the first 10^2 Fibonacci numbers end in 4.
From _Wolfdieter Lang_, Feb 08 2017: (Start)
Curious cubic identities:
3^3 + 7^3 + 0^3 = 370, 33^3 + 67^3 + (00)^3 = 336700, 333^3 + 667^3 + (000)^3 = 333667000, ...
3^3 + 7^3 + 1^3 = 371, 33^3 + 67^3 + (01)^3 = 336701, 333^3 + 667^3 + (001)^3 = 333667001, ...
4^3 + 0^3 + 7^3 = 407, 34^3 + (00)^3 + 67^3 = 340067 , 334^3 + (000)^3 + 677^3 = 334000677, ... (End)

Seiichi Manyama, Table of n, a(n) for n = 0..1001
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A072702, A093137, A109344, A246057, A281858, A281859, A281860.

Cf. A000045, A000326, A001045, A002277, A005188, A073553, A199682.

s = Fibonacci@ Range[10^5]; Table[Count[Take[s, 10^n], m_ /; Mod[m, 10] == 4], {n, 0, Floor@ Log10@ Length@ s}] (* or *) Table[Boole[n > 0] Ceiling[10^n/15], {n, 0, 20}] (* or *) CoefficientList[Series[x (1 - 4 x)/((1 - x) (1 - 10 x)), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *)

a(n)=(10^n+13)\15 \\ Charles R Greathouse IV, Jun 05 2011

a(n) = ceiling((2/30)*10^n) - Benoit Cloitre, Aug 27 2002
From Paul Barry, Mar 24 2004: (Start)
G.f.: x*(1 - 4*x)/((1 - x)*(1 - 10*x)).
a(n) = 10^n/15 + 1/3 for n>0. (End)
a(n) = 10*a(n-1) - 3 for n>1. - Vincenzo Librandi, Dec 07 2010 [Immediate consequence of the previous formula, R. J. Mathar]
From Eric M. Schmidt, Oct 28 2012: (Start)
For n>0, a(n) = A199682(n-1)/3 = (2*10^(n-1) + 1)/3.
For n>=2, a(n+1) = a(n) + 6*10^n. (End)
From Elmo R. Oliveira, Jul 22 2025: (Start)
E.g.f.: (-6 + 5*exp(x) + exp(10*x))/15.
a(n) = 11*a(n-1) - 10*a(n-2) for n >= 3.
a(n) = A073553(n)/2 for n >= 1. (End)

A073552 merged into this sequence by Eric M. Schmidt, Oct 28 2012

cp's OEIS Frontend

A101337 Sum of (each digit of n raised to the power (number of digits in n)).

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A093137 Expansion of (1-7*x)/((1-x)*(1-10*x)).

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A010346 Base-5 Armstrong or narcissistic numbers (written in base 10).

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A010354 Base-8 Armstrong or narcissistic numbers (written in base 10).

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A010344 Base-4 Armstrong or narcissistic numbers (written in base 10).

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A007532 Handsome numbers: sum of positive powers of its digits; a(n) = Sum_{i=1..k} d[i]^e[i] where d[1..k] are the decimal digits of a(n), e[i] > 0.

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Haskell

Maple

Mathematica

Python

Formula

A010348 Base-6 Armstrong or narcissistic numbers (written in base 10).

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A093137 Expansion of (1-7x)/((1-x)(1-10*x)).