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.

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A271977 G_6(n), where G is the Goodstein function defined in A266201.

Original entry on oeis.org

0, 139, 1751, 187243, 16777215, 33554571, 50333399, 84073323, 134217727, 134217867, 134219479, 134404971, 150994943
Offset: 3

Views

Author

Natan Arie Consigli, Apr 24 2016

Keywords

Comments

The next term (line break for better formatting) is a(16) = \
1619239197880733074062994004113160848331305687934176134326809 \
538279709713884753268291640071900343455846003089194770060104834018705547.
a(17) = 2.870...*10^1585, a(18) = 6.943...*10^169099. - Pontus von Brömssen, Sep 24 2020

Examples

			Find G_6(7):
G_1(7) = B_2(7)-1= B_2(2^2+2+1)-1 = 3^3+3+1-1 = 30;
G_2(7) = B_3(G_1(7))-1 = B_3(3^3+3)-1 =  4^4+4-1 = 259;
G_3(7) = B_4(G_2(7))-1 = 5^5+3-1 = 3127;
G_4(7) = B_5(G_3(7))-1 = 6^6+2-1 = 46657;
G_5(7) = B_6(G_4(7))-1 = 7^7+1-1 = 823543;
G_6(7) = B_7(G_5(7))-1 = 8^8-1 = 16777215.
		

Crossrefs

Cf. A056004: G_1(n); A057650: G_2(n); A059934: G_3(n); A059935: G_4(n); A059936: G_5(n); this sequence: G_6(n); A271978: G_7(n); A271979: G_8(n); A271985: G_9(n); A271986: G_10(n); A266201: G_n(n).

Programs

  • Python
    from sympy.ntheory.factor_ import digits
    def bump(n,b):
      s=digits(n,b)[1:]
      l=len(s)
      return sum(s[i]*(b+1)**bump(l-i-1,b) for i in range(l) if s[i])
    def A271977(n):
      if n==3: return 0
      for i in range(2,8):
        n=bump(n,i)-1
      return n # Pontus von Brömssen, Sep 24 2020

Extensions

a(10) corrected by Pontus von Brömssen, Sep 24 2020

A056041 Value for which b(a(n))=0 when b(2)=n and b(k+1) is calculated by writing b(k) in base k, reading this as being written in base k+1 and then subtracting 1.

Original entry on oeis.org

2, 3, 5, 7, 23, 63, 383, 2047
Offset: 0

Views

Author

Henry Bottomley, Aug 04 2000

Keywords

Comments

a(8)=3*2^(3*2^27+27)-1 which is more than 10^(10^8) and equal to the final base of the Goodstein sequence starting with g(2)=4; indeed, apart from the initial term, the sequence starting with b(2)=8 is identical to the Goodstein sequence starting with g(2)=4. The initial terms of a(n) [2, 3, 5 and 7] are equal to the initial terms of the equivalent final bases of Goodstein sequences starting at the same points. a(9)=2^(2^(2^70+70)+2^70+70)-1 which is more than 10^(10^(10^20)).
It appears that if n is even then a(n) is one less than three times a power of two, while if n is odd then a(n) is one less than a power of two.
Comment from John Tromp, Dec 02 2004: The sequence 2,3,5,7,3*2^402653211 - 1, ... gives the final base of the Goodstein sequence starting with n. This is an example of a very rapidly growing function that is total (i.e. defined on any input), although this fact is not provable in first-order Peano Arithmetic. See the links for definitions. This grows even faster than the Friedman sequence described in the Comments to A014221.
In fact there are two related sequences: (i) The Goodstein function l(n) = number of steps for the Goodstein sequence to reach 0 when started with initial term n >= 0: 0, 1, 3, 5, 3*2^402653211 - 3, ...; and (ii) the same sequence + 2: 2, 3, 5, 7, 3*2^402653211 - 1, ..., which is the final base reached. Both grow too rapidly to have their own entries in the database.
Related to the hereditary base sequences - see cross-reference lines.
This sequence gives the final base of the weak Goodstein sequence starting with n; compare A266203, the length of the weak Goodstein sequence. a(n) = A266203(n) + 2.

Examples

			a(3)=7 because starting with b(2)=3=11 base 2, we get b(3)=11-1 base 3=10 base 3=3, b(4)=10-1 base 4=3, b(5)=3-1 base 5=2, b(6)=2-1 base 6=1 and b(7)=1-1 base 7=0.
		

Crossrefs

Equals A266203 + 2.
Steps of strong Goodstein sequences: A056004, A057650, A059934, A059935, A059936, A271977.
Strong Goodstein sequences: A215409, A056193, A266204, A222117, A059933.
Woodall numbers: A003261.

A222112 Initial step in Goodstein sequences: write n-1 in hereditary binary representation, then bump to base 3.

Original entry on oeis.org

0, 1, 3, 4, 27, 28, 30, 31, 81, 82, 84, 85, 108, 109, 111, 112, 7625597484987, 7625597484988, 7625597484990, 7625597484991, 7625597485014, 7625597485015, 7625597485017, 7625597485018, 7625597485068, 7625597485069, 7625597485071, 7625597485072, 7625597485095
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 13 2013

Keywords

Comments

See A056004 for an alternate version.

Examples

			n = 19: 19 - 1 = 18 = 2^4 + 2^1 = 2^2^2 + 2^1
-> a(19) = 3^3^3 + 3^1 = 7625597484990;
n = 20: 20 - 1 = 19 = 2^4 + 2^1 + 2^0 = 2^2^2 + 2^1 + 2^0
-> a(20) = 3^3^3 + 3^1 + 3^0 = 7625597484991;
n = 21: 21 - 1 = 20 = 2^4 + 2^2 = 2^2^2 + 2^2
-> a(21) = 3^3^3 + 3^3 = 7625597485014.
		

References

  • Helmut Schwichtenberg and Stanley S. Wainer, Proofs and Computations, Cambridge University Press, 2012; 4.4.1, page 148ff.

Crossrefs

Cf. A056004: G_1(n), A057650 G_2(n), A056041; A266201: G_n(n);
Cf. A215409: G_n(3), A056193: G_n(4), A266204: G_n(5), A266205: G_n(6), A222117: G_n(15), A059933: G_n(16), A211378: G_n(19).

Programs

  • Haskell
    -- See Link
    
  • PARI
    A222112(n)=sum(i=1, #n=binary(n-1), if(n[i],3^if(#n-i<2, #n-i, A222112(#n-i+1)))) \\ See A266201 for more general code. - M. F. Hasler, Feb 13 2017, edited Feb 19 2017

A271978 G_7(n), where G is the Goodstein function defined in A266201.

Original entry on oeis.org

0, 173, 2454, 332147, 37665879, 774841151, 1162263921, 1937434592, 2749609302, 3486784574, 3486786855, 3487116548, 3524450280
Offset: 3

Views

Author

Natan Arie Consigli, Apr 30 2016

Keywords

Comments

a(16) is too big to include - see b-file. a(17) = 9.221...*10^2347, a(18) = 2.509...*10^316952. - Pontus von Brömssen, Sep 25 2020

Examples

			Find G_7(7):
G_1(7) = B_2(7)-1= B[2](2^2+2+1)-1 = 3^3+3+1-1 = 30;
G_2(7) = B_3(G_1(7))-1 = B[3](3^3+3)-1 =  4^4+4-1 = 259;
G_3(7) = B_4(G_2(7))-1 = 5^5+3-1 = 3127;
G_4(7) = B_5(G_3(7))-1 = 6^6+2-1 = 46657;
G_5(7) = B_6(G_4(7))-1 = 7^7+1-1 = 823543;
G_6(7) = B_7(G_5(7))-1 = 8^8-1 = 16777215;
G_7(7) = B_8(G_6(7))-1 = 7*9^7+7*9^6+7*9^5+7*9^4+7*9^3+7*9^2+7*9+7-1 = 37665879.
		

Crossrefs

Cf. A056004: G_1(n); A057650: G_2(n); A059934: G_3(n); A059935: G_4(n); A059936: G_5(n); A271977: G_6(n); this sequence: G_7(n); A271979: G_8(n); A271985: G_9(n); A271986: G_10(n); A266201: G_n(n).

Programs

  • Python
    from sympy.ntheory.factor_ import digits
    def bump(n,b):
      s=digits(n,b)[1:]
      l=len(s)
      return sum(s[i]*(b+1)**bump(l-i-1,b) for i in range(l) if s[i])
    def A271978(n):
      if n==3: return 0
      for i in range(2,9):
        n=bump(n,i)-1
      return n # Pontus von Brömssen, Sep 25 2020

Extensions

a(9) corrected by Pontus von Brömssen, Sep 25 2020

A271979 G_8(n), where G is the Goodstein function defined in A266201.

Original entry on oeis.org

0, 211, 3325, 555551, 77777775, 20000000211, 30000003325, 50000555551, 70077777775, 100000000211, 100000003325, 100000555551, 100077777775
Offset: 3

Views

Author

Natan Arie Consigli, Apr 30 2016

Keywords

Comments

At least half of the digits of every term (except a(14)) are the same.
Let n > 0:
a(4n) mod 100 = 211;
a(4n+1) mod 1000 = 3325;
a(4n+2) mod 1000000 = 555551;
a(4n+3) mod 100000000 = 77777775;
Proof for a(4n):
If x is divisible by 4 its hereditary representation in base 2 has all summands divisible by 4 and it cannot have the summands 1 and 2.
If we calculate G_1(x) we would end with:
G_1(x) = B_2(x)-1.
Clearly, B_2(x) = 3^a + 3^b + ... is divisible by 3^3 = 27 and that would mean that the representation of B_2(x)-1 would be B_2(x)-1 = X_3 + 2*3^2+2*3+2.
From now on, let X_n be a sum of powers of n (greater than the right term).
We finish proving the statement by calculating G_8(x):
G_2(x) = B_3(X_3 +2*3^2+2*3+2)-1 = X_4 + 2*4^2+2*4+2-1;
G_3(x) = B_4(X_4 +2*4^2+2*4-1)-1 = X_5 + 2*5^2+2*5+1-1;
G_4(x) = B_5(X_5 +2*5^2+2*5)-1 = X_6 + 2*6^2+2*6-1;
G_5(x) = B_6(X_6 +2*6^2+6+5)-1 = X_7 + 2*7^2+7+5-1;
G_6(x) = B_7(X_7 +2*7^2+7+4)-1 = X_8 + 2*8^2+8+4-1;
G_7(x) = B_8(X_8 +2*8^2+8+3)-1 = X_9 + 2*9^2+9+3-1;
G_8(x) = B_9(X_9 +2*9^2+9+2)-1 = X_10 + 2*10^2+10+2-1 = X_10 + 211;
So finally G_8(x) mod 100 = 211.
The other cases can be proved using the same reasoning.
a(17) = 3.3330...*10^3333, a(18) = 5.555550...*10^555555. - Pontus von Brömssen, Sep 25 2020

Examples

			Calculate G_8(5):
G_1(5) = B_2(5)-1 = B_2(2^2+1)-1 = 27;
G_2(5) = B_3(3^3)-1 = 4^4-1 = 255;
G_3(5) = B_4(3*4^3 + 3*4^2 + 3*4 + 3)-1 = 3*5^3 + 3*5^2 + 3*5 + 3-1 = 467;
G_4(5) = B_5(3*5^3 + 3*5^2 + 3*5 + 2)-1 = 3*6^3 + 3*6^2 + 3*6 + 2-1 = 775;
G_5(5) = B_6(3*6^3 + 3*6^2 + 3*6 + 1)-1 = 3*7^3 + 3*7^2 + 3*7 + 1-1 = 1197;
G_6(5) = B_7(3*7^3 + 3*7^2 + 3*7)-1 = 3*8^3 + 3*8^2 + 3*8-1 = 1751;
G_7(5) = B_8(3*8^3 + 3*8^2 + 2*8 + 7)-1 = 3*9^3 + 3*9^2 + 2*9 + 7-1 = 2454;
G_8(5) = B_9(3*9^3 + 3*9^2 + 2*9 + 6)-1 = 3*10^3 + 3*10^2 + 2*10 + 6-1 = 3325.
		

Crossrefs

Cf. A056004: G_1(n); A057650: G_2(n); A059934: G_3(n); A059935: G_4(n); A059936: G_5(n); A271977: G_6(n); A271978: G_7(n); this sequence: G_8(n); A271985: G_9(n); A271986: G_10(n); A266201: G_n(n).

Programs

  • Python
    from sympy.ntheory.factor_ import digits
    def bump(n,b):
      s=digits(n,b)[1:]
      l=len(s)
      return sum(s[i]*(b+1)**bump(l-i-1,b) for i in range(l) if s[i])
    def A271979(n):
      if n==3: return 0
      for i in range(2,10):
        n=bump(n,i)-1
      return n # Pontus von Brömssen, Sep 25 2020

Extensions

Incorrect program and terms removed by Pontus von Brömssen, Sep 25 2020

A271985 G_9(n), where G is the Goodstein function defined in A266201.

Original entry on oeis.org

0, 253, 4382, 885775, 150051213, 570623341475, 855935016215, 1426559238830, 1997331745490, 3138428376974, 3138428381103, 3138429262496, 3138578427934
Offset: 3

Views

Author

Natan Arie Consigli, Apr 30 2016

Keywords

Comments

a(17) = 2.066...*10^4574. - Pontus von Brömssen, Sep 25 2020

Examples

			Compute G_9(10):
G_1(10)= B_2(10)-1 = B_2(2^(2+1)+2)-1 = 3^(3+1)+3-1 = 83;
G_2(10) = B_3(3^(3+1)+2)-1 = 4^(4+1)+2-1 = 1025;
G_3(10) = B_4(4^(4+1)+1)-1 = 5^(5+1)+1-1 = 15625;
G_4(10) = B_5(5*5^(5+1))-1 = 6^(6+1)-1= 279935;
G_5(10) = B_6(5*6^6+5*6^5+5*6^4+5*6^3+5*6^2+5*6+5)-1 = 5*7^7+5*7^5+5*7^4+5*7^3+5*7^2+5*7+5-1 = 4215754;
G_6(10) = B_7(5*7^7+5*7^5+5*7^4+5*7^3+5*7^2+5*7+4)-1 = 5*8^8+5*8^5+5*8^4+5*8^3+5*8^2+5*8+4-1 = 84073323;
G_7(10) = B_8(5*8^8+5*8^5+5*8^4+5*8^3+5*8^2+5*8+3)-1 = 5*9^9+5*9^5+5*9^4+5*9^3+5*9^2+5*9+3-1 = 1937434592;
G_8(10) = B_9(5*9^9+5*9^5+5*9^4+5*9^3+5*9^2+5*9+2)-1 = 5*10^10+5*10^5+5*10^4+5*10^3+5*10^2+5*10+2-1 = 50000555551;
G_9(10) = B_10(5*10^10+5*10^5+5*10^4+5*10^3+5*10^2+5*10+1)-1 = 5*11^11+5*11^5+5*11^4+5*11^3+5*11^2+5*11+1-1 = 1426559238830.
		

Crossrefs

Cf. A056004: G_1(n); A057650: G_2(n); A059934: G_3(n); A059935: G_4(n); A059936: G_5(n); A271977: G_6(n); A271978: G_7(n); A271979: G_8(n); this sequence: G_9(n); A271986: G_10(n); A266201: G_n(n).

Programs

  • Python
    from sympy.ntheory.factor_ import digits
    def bump(n,b):
      s=digits(n,b)[1:]
      l=len(s)
      return sum(s[i]*(b+1)**bump(l-i-1,b) for i in range(l) if s[i])
    def A271985(n):
      if n==3: return 0
      for i in range(2,11):
        n=bump(n,i)-1
      return n # Pontus von Brömssen, Sep 25 2020

Extensions

Incorrect program and terms removed by Pontus von Brömssen, Sep 25 2020

A271986 G_10(n), where G is the Goodstein function defined in A266201.

Original entry on oeis.org

0, 299, 5643, 1357259, 273624711, 17832200896811, 26748301350411, 44580503598539, 62412976762503, 106993205379371, 106993205384715, 106993206736331, 106993479003783
Offset: 3

Views

Author

Natan Arie Consigli, May 01 2016

Keywords

Comments

a(17) = 1.926...*10^6103. - Pontus von Brömssen, Sep 25 2020

Crossrefs

Cf. A056004: G_1(n); A057650: G_2(n); A059934: G_3(n); A059935: G_4(n); A059936: G_5(n); A271977: G_6(n); A271978: G_7(n); A271979: G_8(n); A271985: G_9(n); this sequence: G_10(n); A266201: G_n(n).

Programs

  • Python
    from sympy.ntheory.factor_ import digits
    def bump(n,b):
      s=digits(n,b)[1:]
      l=len(s)
      return sum(s[i]*(b+1)**bump(l-i-1,b) for i in range(l) if s[i])
    def A271986(n):
      if n==3: return 0
      for i in range(2,12):
        n=bump(n,i)-1
      return n # Pontus von Brömssen, Sep 25 2020

Extensions

Incorrect program and terms removed by Pontus von Brömssen, Sep 25 2020

A273004 Sum of coefficients in the hereditary representation of n in base 2.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 4, 5, 6, 7, 7, 8, 9, 10, 8, 9, 10, 11, 11, 12, 13, 14, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14, 15, 9, 10, 11, 12, 12, 13, 14, 15, 13, 14, 15, 16, 16, 17, 18, 19, 6, 7, 8, 9, 9, 10, 11, 12, 10, 11, 12, 13, 13, 14, 15, 16, 10, 11, 12, 13, 13, 14, 15, 16, 14, 15, 16, 17, 17, 18, 19, 20, 11, 12, 13, 14, 14
Offset: 0

Views

Author

M. F. Hasler, May 12 2016

Keywords

Comments

The hereditary representation of a number n in base b is a [possibly empty] sum (possibly represented as a list) of monomials of the form m*b^e (possibly represented as a list [m,e] or as a single number m if e = 0) with coefficients 0 < m < b, and the (strictly increasing) exponents e > 0 recursively again expressed in the same form. Thus 0 = [], 1 = 1*b^0 = [1], b = 1*b^1 = [[1, [1]]] etc.

Examples

			266 = 1*2^1 + 1*2^(1+1*2^1) + 1*2^(1*2^(1+1*2^1)) which can be represented as [[1, [1]], [1, [1, [1, [1]]]], [1, [[1, [1, [1, [1]]]]]]], and there are 11 "1"s, therefore a(266) = 11.
		

Crossrefs

Cf. A056004, A222112, A273005 (base 10 analog), A033922, A361838.

Programs

  • Mathematica
    a[n_] := a[n] = Total[1 + a /@ Log2[DeleteCases[NumberExpand[n, 2], 0]]]; (* Vladimir Reshetnikov, Dec 21 2023 *)
  • PARI
    (hr(n,b=2)=if(1<#n=digits(n,b),my(v=if(n[#n],[n[#n]],[]));forstep(i=#n-1,1,-1,n[i]&&v=concat(v,[[n[i],hr(#n-i,b)]]));v,n));(cc(v)=if(type(v)=="t_VEC",sum(i=1,#v,cc(v[i])),v)); a(n)=cc(hr(n))
    
  • Python
    def A273004(n):
      s=format(n,'b')[::-1]
      return sum(1+A273004(i) for i in range(len(s)) if s[i]=='1') # Pontus von Brömssen, Sep 17 2020

Formula

If n = Sum_{j=1..k} 2^e_j where 0 <= e_1 < ... < e_k, then a(n) = k + Sum_{j=1..k} a(e_j). - Pontus von Brömssen, Sep 17 2020
a(n) = A033922(n) + A361838(floor(n/2)) for n > 1. - Andrei Zabolotskii, Aug 24 2025

A273005 Sum of coefficients in the hereditary representation of n in base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 3
Offset: 0

Views

Author

M. F. Hasler, May 12 2016

Keywords

Examples

			266 = 6 + 6*10^1 + 2*10^2 which can be represented as [6, [6, [1]], [2, [2]]], therefore a(266) = 6 + 6 + 1 + 2 + 2 = 17.
		

Crossrefs

Programs

  • PARI
    (hr(n,b=10)=if(1<#n=digits(n,b),my(v=if(n[#n],[n[#n]],[]));forstep(i=#n-1,1,-1,n[i]&&v=concat(v,[[n[i],hr(#n-i,b)]]));v,n));(cc(v)=if(type(v)=="t_VEC",sum(i=1,#v,cc(v[i])),v)); a(n)=cc(hr(n,10))
    
  • Python
    def A273005(n):
      s=str(n)[::-1]
      return sum(int(s[i])+A273005(i) for i in range(len(s)) if s[i]!='0') # Pontus von Brömssen, Sep 17 2020

Formula

If n = Sum_{j=1..k} d_j*10^(e_j) where 0 <= e_1 < ... < e_k and 1 <= d_j <= 9, then a(n) = Sum_{j=1..k} (d_j + a(e_j)). - Pontus von Brömssen, Sep 17 2020

A296441 Array A(n, k) = G_k(n) where G_k(n) is the k-th term of the Goodstein sequence of n, read by antidiagonals.

Original entry on oeis.org

0, 0, 1, 0, 0, 2, 0, 0, 2, 3, 0, 0, 1, 3, 4, 0, 0, 0, 3, 26, 5, 0, 0, 0, 2, 41, 27, 6, 0, 0, 0, 1, 60, 255, 29, 7, 0, 0, 0, 0, 83, 467, 257, 30, 8, 0, 0, 0, 0, 109, 775, 3125, 259, 80, 9, 0, 0, 0, 0, 139, 1197, 46655, 3127, 553, 81, 10, 0, 0, 0, 0, 173, 1751, 98039, 46657, 6310, 1023, 83, 11
Offset: 0

Views

Author

Iain Fox, Dec 12 2017

Keywords

Comments

G_0(n) = n. To get to the second term in the row, convert n to hereditary base 2 representation (see links), replace each 2 with a 3, and subtract 1. For the third term, convert the second term (G_1(n)) into hereditary base 3 notation, replace each 3 with a 4, and subtract one. This pattern continues until the sequence converges to 0, which, by Goodstein's Theorem, occurs for all n.

Examples

			| n\k |  0   1    2     3      4      5       6       7       8       9  ...
|-----|------------------------------------------------------------------------
|  0  |  0,  0,   0,    0,     0,     0,      0,      0,      0,      0, ...
|  1  |  1,  0,   0,    0,     0,     0,      0,      0,      0,      0, ...
|  2  |  2,  2,   1,    0,     0,     0,      0,      0,      0,      0, ...
|  3  |  3,  3,   3,    2,     1,     0,      0,      0,      0,      0, ...
|  4  |  4, 26,  41,   60,    83,   109,    139,    173,    211,    253, ...
|  5  |  5, 27, 255,  467,   775,  1197,   1751,   2454,   3325,   4382, ...
|  6  |  6, 29, 257, 3125, 46655, 98039, 187243, 332147, 555551, 885775, ...
| ... |
		

Crossrefs

n-th row: A000004 (n=0), A000007 (n=1), A215409 (n=3), A056193 (n=4), A266204 (n=5), A266205 (n=6), A271554 (n=7), A271555 (n=8), A271556 (n=9), A271557 (n=10), A271558 (n=11), A271559 (n=12), A271560 (n=13), A271561 (n=14), A222117 (n=15), A059933 (n=16), A271562 (n=17), A271975 (n=18) A211378 (n=19), A271976 (n=20).
k-th column: A001477 (k=0), A056004 (k=1), A057650 (k=2), A059934 (k=3), A059935 (k=4), A059936 (k=5), A271977 (k=6), A271978 (k=7), A271979 (k=8), A271985 (k=9), A271986 (k=10).
G_n(n) = A266201(n) (main diagonal of array).

Programs

  • PARI
    B(n, b)=sum(i=1, #n=digits(n, b), n[i]*(b+1)^if(#n
    				
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