

A187813


Numbers n whose baseb digit sum is not b for all bases b >= 2.


19



0, 1, 2, 4, 8, 14, 30, 32, 38, 42, 44, 54, 60, 62, 74, 84, 90, 98, 102, 104, 108, 110, 114, 128, 138, 140, 150, 152, 158, 164, 168, 174, 180, 182, 194, 198, 200, 212, 224, 228, 230, 234, 240, 242, 252, 270, 278, 282, 284, 294, 308, 312, 314, 318, 332, 338, 348
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,3


COMMENTS

Except for 1, every number is even.
No number ends in 6.
Numbers neither in A018900 nor in A226636 nor in A226969 nor in A227062 nor in A227080 nor ... .  R. J. Mathar, Sep 02 2013
From Hieronymus Fischer, Mar 27 2014, May 09 2014: (Start)
A079696 and this sequence have no terms in common.
Numbers which satisfy m == 1 (mod j) and m > j^2 for any j > 1 are not terms.
Example 1: m = 10^k, k>1, is not a term since 10^k == 1 (mod 9) and 10^k > 9^2.
Example 2: m = 1 + 3k, k > 3, is not a term, since m > 3(1+3) > 3^2.
This is the complement of the disjunction of A079696 with A239708.
Disregarding the first 3 terms, these are the numbers which are in A008864 but not in A239708. This leads to the following characterization: A number m > 2 is a term, i.e., satisfies digitalSum_b(m) <> b for all b > 1, if and only m is a prime number + 1 and m is not the sum of two distinct powers of 2.
a(6) is the only term such that a(n) = Prime(n) + 1. For n < 6, we have a(n) < Prime(n) + 1, and for n > 6, we have a(n) > Prime(n) + 1.
(End)


LINKS

Hieronymus Fischer, Table of n, a(n) for n = 1..10000


FORMULA

From Hieronymus Fischer, Mar 27 2014: (Start)
A239703(a(n)) = 0.
a(n+1) = min (p > a(n)  A239703(p) = 0)
[for a Smalltalk implementation see Prog section, method A187813NextTerm version 1].
a(n+1) = 1 + min (p > a(n)  p is prime AND ((q := p+1  2^floor(log_2(p+1)) = 0) OR (2^floor(log_2(q)) <> q)))
[for a Smalltalk implementation see Prog section, method A187813NextTerm version 2].
a(n) > Prime(n), for n > 5.
a(n  m) < Prime(n), for n > 1, where m := i*(i1)/2 + j  1, i := floor(log_2(Prime(n))), j := floor(log_2(Prime(n)  2^i)).
a(n  m) < Prime(n), for n > 32, where m := i*(i1)/2 + j  16 with i and j above.
a(n) = Prime(n + m  3) + 1, where m = max ( k  A239708(k) < a(n)), n > 3.
Remark: This identity can be used to calculate a(n) recursively. For a Smalltalk implementation see Prog section, methods A187813rec and A187813With: estimate.
With same conditions: a(n) = A008864(n + m  3).
a(n  m + 3) = Prime(n) + 1, where m = max ( k  A239708(k) < Prime(n)), n > 3, provided Prime(n) + 1 is not a term of A239708.
(End)


EXAMPLE

8 has binary expansion (1,0,0,0) whose digit sum 1 is not 2,
ternary expansion (2,2) whose digit sum 4 is not 3,
quaternary expansion (2,0) whose digit sum 2 is not 4,
5ary expansion (1,3) whose digit sum 4 is not 5,
6ary expansion (1,2) whose digit sum 3 is not 6,
7ary expansion (1,1) whose digit sum 2 is not 7,
8ary expansion (1,0) whose digit sum 1 is not 8,
and bary expansion (8) when b>8 whose digit sum is 8 not b. Therefore, 8 is in the sequence.
3 has binary expansion (1,1) whose digit sum is 2, so 3 is not in the sequence.
From Hieronymus Fischer, Apr 10 2014: (Start)
a(10) = 42 (the 13th prime + 1)
a(100) = 618 (the 113th prime + 1)
a(1000) = 8172 (the 1026th prime + 1)
a(10^4) = 105254 (the 10042nd prime + 1)
a(10^5) = 1300464 (the 100056th prime + 1)
a(10^6) = 15486872 (the 1000063th prime + 1)
a(10^7) = 179425944 (the 10000071st prime + 1)
a(10^8) = 2038076324 (the 10^8 +84th prime + 1)
a(10^9) = 22801765334 (the 10^9 +92nd prime + 1)
a(10^10) = 252097803264 (the 10^10 +102nd prime + 1)
[calculation for large numbers processed with Smalltalk method A187813With: estimate; see Prog section]
(End)


PROG

(Sage)
n=1000 #change n for more terms
S=[]
for i in [0..n]:
test=False
for b in [2..i]:
if sum(Integer(i).digits(base=b))==b:
test=True
break
if not test:
S.append(i)
S
# From Hieronymus Fischer, Apr 10 2014: (Start)
(Smalltalk)
A187813NextTerm
"Calculates the next term of A187813 greater than the receiver, i.e., calculates a(n+1) from a(n).
Usage: a(n) A187813NextTerm
Answer: a(n+1)
Version 1: Using numOfBasesWithDigitalSumEQBase from A239703 ==> fast calculation, since only the divisors of <an> have to tested to be candidates for bases b with baseb digital sum equal to b"
 an 
an := self + 1.
[an numOfBasesWithDigitalSumEQBase > 0]
whileTrue: [an := an+1].
^an

A187813NextTerm
"Calculates the next term of A187813 greater than the receiver, i.e., calculates a(n+1) from a(n).
Usage: a(n) A187813NextTerm
Answer: a(n+1)
Version 2: Using the equivalence with A008864 and A239708 ==> even much more faster calculation"
 p q 
self < 0 ifTrue: [^0].
self = 0 ifTrue: [^1].
self = 1 ifTrue: [^2].
p := (self  1) nextPrime.
q := p+1(2 raisedToInteger: (p+1 integerFloorLog: 2)).
[q > 0 and: [(2 raisedToInteger: (q integerFloorLog: 2))  q = 0]] whileTrue: [p := p nextPrime.
q := p + 1  (2 raisedToInteger: (p + 1 integerFloorLog: 2))].
^p + 1

A187813
"Calculates the nth term of A187813, iteratively.
Usage: n A187813
Answer: a(n)"
 an n 
n := self.
n < 3 ifTrue: [^#(0 1) at: n].
an := 2.
4 to: n do: [:i an := an A187813NextTerm].
^an

A187813rec
"Calculates the nth term of A187813, using the recursive method <A187813With: param>
Usage: n A187813
Answer: a(n)"
self < 3 ifTrue: [^#(0 1) at: self].
^self A187813With: self prime

A187813With: estimate
"Method to calculate the nth term of A187813 based on the value estimate, recursively. The nth prime is a adequate estimate. Valid for n > 2.
Usage: n A187813With: estimate
Answer: a(n)"
 x m 
(x:=((m:= estimate A239708inv)+self3) prime + 1) = estimate
ifFalse: [^self A187813With: x].
(m + 1) A239708 = x
ifTrue: [^self A187813With: x + 4].
^x
[End]


CROSSREFS

Cf. A018900, A052224.
Cf. A007953, A079696, A008864, A239708, A000040.
Cf. A239703, A239705.
Sequence in context: A283835 A244933 A118560 * A038024 A061297 A130711
Adjacent sequences: A187810 A187811 A187812 * A187814 A187815 A187816


KEYWORD

nonn,base


AUTHOR

Tom Edgar, Aug 30 2013


STATUS

approved



